NCERT Solutions Class 11 Biology

Biology is the most significant subject for students who want to pursue a career in medicine. NEET is the single entrance exam for medical colleges across India. This examination’s syllabus comprises both Class 11 Biology and Class 12 Biology NCERT books. In several Indian schools, NCERT Books for Class 11 Biology have been recommended. All concepts and topics are described in great detail and clarity in these books. As a result, students should refer to NCERT textbooks to prepare better for the exam.

NCERT Solutions for Class 11 Biology

CBSE Class 11 Biology covers a wide range of key topics such as the living world, biological taxonomy, structural organisation in plants and animals, the cell, cell division, biomolecules, plant physiology, human physiology, and so on. A thorough understanding of the fundamentals of CBSE Class 11 Biology Notes is vital. These chapters’ themes are essential in higher classes to readily understand difficult topics. 

Those aiming to secure good marks should look at remembering and understanding these principles. Particularly in class 11, because this will serve as the foundation for all future learning. Hence, students should extensively cover each topic mentioned in the NCERT Textbooks and be confident in answering any question from it. In addition, the Biology Books for Class 11 feature a number of questions and activities at the end of each chapter. These NCERT Solutions for Class 11 Biology come to the rescue if they get stuck in the middle.

NCERT Class 11 Biology Chapter wise Solutions

The Class 11 Biology Solutions are organised chapter-by-chapter and question-by-question to make it simple for students to access and utilise. The NCERT textbook for Class 11 Biology contains 22 chapters. Students can download the chapter-wise solutions by clicking on the title of the chapter from the list given below. It is advisable to first attempt these questions on your own and then compare their answers to these solutions to see whether they are correct or not. 

Biology NCERT Solutions for Class 11 

NCERT Solutions for Class 11 Biology are useful not only for Class 11 exams, but also for entrance tests such as NEET, JEE, medical entrance exams, and other competitive exams. These solutions were developed to assist students in grasping the most knowledge with the least amount of mental effort. Some questions given in the NCERT Book are tricky and difficult to understand. The most common problems faced by students are difficult terms and procedures. For this, students may download the chapter-wise solutions Class 11 Biology Solutions which also contain question-wise answers. These answers are explained in great detail and in simple language. You can read these solutions and prepare your own notes for each unique question.

NCERT Solution for Class 11 Biology

Extramarks has created these solutions for the questions found at the end of each chapter in the Class 11 Biology NCERT book. These solutions were created by subject matter experts who have decades of experience in the education system to ensure that you can understand the solutions in a step-by-step manner. When using them for homework, you can present your answers in a way that helps you get good grades in your classes.

We have ensured that none of these options will cost you anything. You don’t even have to read these NCERT Solutions for Class 11 Biology online because we’ve included Class 11 download links that you can just click to download and read at your leisure.

Chapter Details & Topics Covered in NCERT Solutions for Class 11 Biology

Chapter 1: The Living World 

This chapter introduces students to a wide range of natural flora and fauna. It begins by asking, “What is living?” before moving on to discuss taxonomic categories and tools. Students become acquainted with the critical subject of genus and species categorisation. They are brought into the fascinating realm of living beings, where they uncover previously discovered knowledge about their environment and habits. They are also educated about biodiversity and the importance of preserving it.

Topics included in Chapter 1: The Living World

What is living? Biodiversity; Need for classification; three domains of life; concept of species and taxonomic hierarchy; binomial nomenclature.

Chapter 2: Biological Classification

The chapter contains characteristics of the Whittaker 5 system classification   Kingdoms Monera, Protista, and Fungi. The Kingdoms Plantae and Animalia, also known as the plant and animal kingdoms, will be discussed further in Chapters 3 and 4, respectively. The kingdoms Monera, Protista, Fungi, Plantae, Animalia, Viruses, Viroids, and Lichens will also be covered.

Topics included in Chapter 2: Biological Classification

Five kingdom classification; Salient features and classification of Monera, Protista and Fungi into major groups; Lichens, Viruses and Viroids.

Chapter 3: Plant Kingdom

From this chapter , the students delve into the specifics of the world as it has been classified thus far. They learn about the plant kingdom’s additional classifications in order to make it easier to study. They are exposed to new terms such as Gymnosperms, Angiosperms, and others throughout this chapter. They also learn about the plant’s life cycle and the generational cycle. They have a thorough understanding of the habitat, features, and physiology of plants, which are complex organisms.

Topics included in Chapter 3: Plant Kingdom

Salient features and classification of plants into major groups – Algae, Bryophyta, Pteridophyta and Gymnosperms. (salient and distinguishing features and a few examples of each category).

Chapter 4: Animal Kingdom

Kingdom Animalia is the most important kingdom in this chapter. Porifera, Coelenterata (Cnidaria), Ctenophora, Platyhelminthes, Aschelminthes, Annelida, Arthropoda, Mollusca, Echinodermata, Hemichordata, and Chordata are the phyla into which animals are classified. Animal survival strategies, physiology, environment, and anatomy are also extensively discussed . The first unit, Diversity in Living Organisms, comes to a close with this chapter. It carries a 14 percent weightage in the NEET test and 7 marks in the final exam.

Topics included in Chapter 4: Animal Kingdom

Salient features and classification of animals, non-chordates up to phyla level and chordates up to class level (salient features and distinguishing features of a few examples of each category). (No live animals or specimens should be displayed.)

Chapter 5: Morphology of Flowering Plants

In this chapter, students will learn about plant morphology. It introduces them to the terms and definitions that they must know in order to comply with the requirements. They learn about the roots, its areas, alterations, and so on. They also learn about flowers and stalks.

Topics included in Chapter 5: Morphology of Flowering Plants

Morphology of inflorescence and flower, Description of 01 family: Solanaceae or Liliaceae (to be dealt along with the relevant experiments of the Practical Syllabus).

Chapter 6: Anatomy of Flowering Plants

Students will learn about the internal structure and functional organisation of plants in this chapter. Plant tissue is divided into two types: meristematic and persistent. The basic activities of plant tissue include food assimilation and storage, water and mineral movement, and so on. They’ll also learn about plant tissue composition and types.

Topics included in Chapter 6: Anatomy of Flowering Plants

The Tissues, The Tissue System, Anatomy of Dicotyledonous and Monocotyledonous Plants, Secondary Growth

Chapter 7: Structural Organisation in Animals

It covers multicellular and acellular animal species, as well as structured tissue and its functions. Students have already studied a wide range of species, both unicellular and multicellular.

Earthworms, Cockroaches, and Frogs’ morphology and anatomy are studied in depth by the learner. Structural Organisation in Plants and Animals, has a 5% weighting in NEET and 12 marks in the final exams. This also marks the end of Unit 2.

Topics included in Chapter 7: Structural Organisation in Animals

Animal tissues.

Chapter 8: Cell-the Unit of Life

Cell theory underpins all of biology. Prokaryotic and Eukaryotic cells are both introduced to them. Students study membrane-bound and non-membrane-bound organelles in cells. With the help of informative graphics, they learn about their structure and function as well. Students also learn about other biologists’ contributions to this discipline over time.

Topics included in Chapter 8: Cell the Unit of Life

Cell theory and cell as the basic unit of life, structure of prokaryotic and eukaryotic cells; Plant cell and animal cell; cell envelope; cell membrane, cell wall; cell organelles – structure and function; endomembrane system, endoplasmic reticulum, golgi bodies, lysosomes, vacuoles, mitochondria, ribosomes, plastids, microbodies; cytoskeleton, cilia, flagella, centrioles (ultrastructure and function); nucleus.

Chapter 9: Biomolecules

The student will learn about biomolecules, their functions, and structures, as well as why they are significant, in this chapter. Metabolites, bio-macromolecules such as proteins, and enzymes will be covered. They’ll learn how enzymes play a crucial role in biochemical reactions.

Topics included in Chapter 9: Biomolecules

Chemical constituents of living cells: biomolecules, structure and function of proteins, carbohydrates, lipids, nucleic acids; Enzymes- types, properties, enzyme action.

Chapter 10: Cell Cycle and Cell Division

The principles of the cell cycle and cell division are introduced in this chapter. Students understand the differences and importance of meiosis and mitosis. Interphase, Prophase, Metaphase, Anaphase, Telophase, and Cytokinesis are all phases of the cell cycle that they study . Unit 3: Cell Structure and Functions is completed in this chapter. It carries a 9% weightage in NEET and 15 marks in the final exam.

Topics included in Chapter 10: Cell Cycle and Cell Division

Cell cycle, mitosis, meiosis and their significance

Chapter 11: Transport in Plants

A vascular system in higher plants, including xylem and phloem, is responsible for translocation. Phloem is in charge of transporting food sugar from the source to the sink. This chapter teaches the learner about many parts of the transportation system in plants involving Xylem and Phloem such as imbibing, plasmolysis, osmosis, active transport, and transpiration. Means of Transportation, Plant-Water Relations, Long-Distance Water Transport, Transpiration, Mineral Nutrient Uptake and Transport, and other topics are covered in this chapter. The phloem translocation is explained by the pressure-flow theory.

Topics included in Chapter 11: Transport in Plants

Means of Transport, Plant-Water Relations, Long Distance Transport of Water, Transpiration, Uptake and Transport of Mineral Nutrients, Phloem, Transport: Flow from Source to Sink. Plants obtain a variety of inorganic elements (ions) and salts from their surroundings, especially from water and soil.

Chapter 12: Mineral Nutrition

Students will learn how to identify the various minerals required for plant growth and function in this chapter. They study what qualifies a mineral as essential, as well as its mode of transport and process of absorption.

It also talks about the deficiency symptoms associated with a lack of said minerals, for a more comprehensive understanding of the subject. They are also given a basic overview of nitrogen fixation and its importance.

Topics included in Chapter 12: Mineral Nutrition

Route of transport and absorption, symptoms of mineral deficiency, nitrogen fixation and its importance.

Chapter 13: Photosynthesis in Higher Plants

The photosynthetic mechanism is explained in detail in this chapter. Students learn about the numerous processes, such as the light reaction, the electron transport chain, and the C4 cycle. They also read about photo-phosphorylation, both cyclic and non-cyclic. NADPH and ATP are highlighted in this chapter.

Topics included in Chapter 13: Photosynthesis in Higher Plants

Photosynthesis as a means of autotrophic nutrition; site of photosynthesis, pigments involved in photosynthesis (elementary idea); photochemical and biosynthetic phases of photosynthesis; cyclic and non-cyclic photophosphorylation; chemiosmotic hypothesis; photorespiration; C3 and C4 pathways; factors affecting photosynthesis.

Chapter 14: Respiration in Plants

This chapter regulates cell breath, or the component of the breakdown of food materials within the cell to give energy, as well as the capture of this energy for the synthesis of ATP. This chapter further elaborates on subjects, such as “Do Plants Breathe?” The topic covers the principles of Glycolysis, Aerobic Respiration, The Respiratory Balance Sheet, Fermentation, Amphibolic Pathway, and Respiratory Quotient.

Topics included in Chapter 14: Respiration in Plants

Exchange of gases; cellular respiration – glycolysis, fermentation (anaerobic), TCA cycle and electron transport system (aerobic); energy relations – number of ATP molecules generated; amphibolic pathways; respiratory quotient.

Chapter 15: Plant Growth and Development

Students will learn about the different internal and external elements that influence plant growth and development. They study plant growth stages, growth regulators, differentiation, dedifferentiation, redifferentiation, vernalisation, seed dormancy, and other topics. Plant hormones’ role in plant growth and development is also studied. Unit 4, Plant Physiology, is completed in this chapter. It is worth 6% of NEET points and 18 points in the final test.

Topics included in Chapter 15: Plant Growth and Development

Growth regulators – auxin, gibberellin, cytokinin, ethylene, ABA.

Chapter 16: Digestion and Absorption

In this chapter, students begin to investigate human physiology and get an understanding of how their own bodies work. This chapter is largely concerned with digestion and absorption of digested food. Digestive system problems are briefly discussed as well. The chapter also discusses the many food components that are needed in our daily diet and the problems that arise when they are deficient.

Topics included in Chapter 16: Digestion and Absorption

Human physiology, food digestion, absorption, digestive system disorders, nutritional components and deficits in diet.

Chapter 17: Breathing and Exchange of Gases

Through this chapter, students will understand the idea of breathing and its difference from respiration. The respiratory system and the organs that comprise it are the focus of this study. The exchange and movement of gases such as oxygen and carbon dioxide are thoroughly covered. Students are also given a brief overview of respiratory issues and the situations that lead to them.

Topics included in Chapter 17: Breathing and Exchange of Gases

Respiratory organs in animals (recall only); Respiratory system in humans; mechanism of breathing and its regulation in humans – exchange of gases, transport of gases and regulation of respiration, respiratory volume; disorders related to respiration – asthma, emphysema, occupational respiratory disorders.

Chapter 18: Body Fluids and Circulation

Humans have a sophisticated circulatory system. This subject is covered in this chapter. The learners study the circulatory routes in people. They learn about blood and its components, lymph (tissue fluid), blood types, and so on. Heart diseases as well as electrocardiographs are also covered.

Topics included in Chapter 18: Body Fluids and Circulation

Composition of blood, blood groups, coagulation of blood; composition of lymph and its function; human circulatory system – Structure of human heart and blood vessels; cardiac cycle, cardiac output, ECG; double circulation; regulation of cardiac activity; disorders of circulatory system – hypertension, coronary artery disease, angina pectoris, heart failure.

Chapter 19: Excretory Products and their Elimination

This chapter is about the human excretory system. Students learn about the excretory system’s various components, including the auxiliary organs. They research the process of urine production and kidney function , as well as a variety of renal illnesses.

Topics included in Chapter 19: Excretory Products and their Elimination

Modes of excretion – ammonotelism, ureotelism, uricotelism; human excretory system – structure and function; urine formation, osmoregulation; regulation of kidney function – renin – angiotensin, atrial natriuretic factor, ADH and diabetes insipidus; role of other organs in excretion; disorders – uremia, renal failure, renal calculi, nephritis; dialysis and artificial kidney, kidney transplant.

Chapter 20: Locomotion and Movement

This chapter introduces students to several types of locomotion and movement in both unicellular and multicellular creatures. The skeletal and muscular systems are thoroughly explored. This chapter investigates the numerous voluntary and involuntary muscular motions, muscle anatomy and function, and muscle illnesses such as gout, osteoporosis, and arthritis.

Topics included in Chapter 20: Locomotion and Movement

Skeletal muscle, contractile proteins and muscle contraction.

Chapter 21: Neural Control and Coordination

This chapter talks about the human central and peripheral nerve systems. Students  will study  impulse creation and transmission, as well as synapses and reflex action. They discover that the neuron is the basic building block of the neurological system. With the help of clear pictures, they can also understand the mechanics of seeing and hearing.

Topics included in Chapter 21: Neural Control and Coordination

Neuron and nerves; Nervous system in humans – central nervous system; peripheral nervous system and visceral nervous system; generation and conduction of nerve impulse.

Chapter 22: Chemical Coordination and Integration

This chapter is about the endocrine and exocrine glands that exist in organisms. The human endocrine system, the hormones secreted by it, the functions of said hormones, and the disorders caused by excess or deficiency of the same are given special attention. Students study the mechanism of hormone action as well as the feedback mechanism that governs hormone production. Unit 5, Human Physiology, is completed with this chapter. It is worth 20% of your NEET score and 18 points in the final test.

Topics included in Chapter 22: Chemical Coordination and Integration

Endocrine glands and hormones; human endocrine system – hypothalamus, pituitary, pineal, thyroid, parathyroid, adrenal, pancreas, gonads; mechanism of hormone action (elementary idea); role of hormones as messengers and regulators, hypo – and hyperactivity and related disorders; dwarfism, acromegaly, cretinism, goitre, exophthalmic goitre, diabetes, Addison’s disease.

Note: Diseases related to all the human physiological systems to be taught in brief.

Tips for Acing the NCERT Biology Exam in Class 11

Solutions for Class 11 Biology Solutions are a vital requirement for acing the NCERT Biology Exam in Class 11. So, every student should have CBSE NCERT solutions for class 11 biology on hand for better CBSE exam preparation. The main thing is that your knowledge of the topics in the biology chapters will improve, and you will be able to perform better in tests. Here are some tips for students to follow:-

  • Make a study plan and follow it consistently and diligently.

When studying for the exam, it is critical to keep to a well-planned timeframe and study strategy. To ensure a seamless and sound preparation, students must develop a study strategy after properly analysing the new exam pattern, curriculum, and marking system. They should devote equal time to each of the exam’s main content, with ample study breaks in between. Extra time should be set aside for important and challenging topics, as well as problem-solving activities, practise papers, and review sessions.

  • Get started on your preparations as soon as possible.

Students should begin studying for each topic as soon as possible. Postponing your study schedule will only add pressure and an incomplete study. If you start early enough, you can focus on every detail  and have a more holistic coverage of the curriculum. You can use the NCERT solutions for Class 11 to prepare your own notes based on your understanding. As a consequence, you’ll be able to grasp concepts more quickly and  easily. 

  • Understand the Biology paper’s structure and curriculum in depth.

The CBSE publishes and updates the Class 11 Biology exam syllabus and exam pattern on a regular basis. The questions for the Class 11 Biology examination are all selected from the same collection of topics. Aside from the curriculum, the design of the paper is crucial because it may aid you in developing your own strategy for successfully completing the question paper. Therefore, students should spend their time assessing and solving previous years’ as well as sample question papers to learn more about the structure of the paper and prepare accordingly.

  • Examine a number of different question variations.

Students can practise on a variety of sample papers and CBSE Question Papers For Class 11 Biology. Solving these papers will help you become more familiar with a range of question types. It will also help you have a better understanding of the question structure and the best way to answer a question.

  • Important Terms and Diagrams Should Not Be Ignored

There are a few fundamental concepts in Biology that are essential for exams. In addition to studying and memorising crucial concepts, the topic contains several important illustrations, clearly explained in the NCERT Solutions for Class 11 Biology. Practice the pictures and their labelling thoroughly in order to deliver the proper solution in the exam. It will help you recall and quickly review them.

  • Examine the Biology Syllabus at least a Few Times

Students must prioritise revision sessions in order to recall and retain material for a longer period of time. For the Biology final exam, re-read the themes of each Biology chapter and double-check the marks each chapter carries.

2022-23 Marking Scheme for CBSE Biology Class 11

The CBSE Class 12 syllabus is divided into two terms: Term 1 and Term 2. Subjective/open-ended short answer type/situation-based/case-based questions are included in the CBSE Class 12 Exam Pattern Term 2. The majority of the questions in both Term 1 and Term 2 are objective in nature.

Term – I 

Units Term – I Marks
I Diversity of Living Organisms: Chapter – 1, 2, 3, and 4 15
II Structural Organisation in Plants and Animals: Chapter 5 and 7 08
III Cell: Structure and Function: Chapter – 8 and 9 12

Term – II 

Units Term – II Marks
IV Cell: Structure and Function: Chapter – 10  05
V Plant Physiology: Chapter – 13, 14, and 15 12
VI Human Physiology: Chapter – 17, 18, 19, 20, 21, and 22 18
Total Theory (Term – I and Term – II)  70
Practical Term – I  15
Practical Term – II  15
Total 100

NCERT Solution of Biology Class 11 – Free Download

Extramarks offers free downloads of the most recent solutions developed with the latest question papers and techniques. All of the chapters in the NCERT Biology textbook for Class XI are covered in these solutions. Here you’ll find all of the in-text and chapter-end practice questions, as well as detailed answers to each one. Extramarks’ qualified professors who have substantial expertise in their respective disciplines have developed all of the Class 11 Biology NCERT Solutions on this page.

NCERT Solutions of Biology Class 11

In most cases, students just read an idea and attempt to memorise it. You may not even understand the subject, but you memorise it because it is an important portion of your exam. This is where these solutions come in – the prescribed responses are recreated into a language that is simple to understand and remember. A team of highly experienced professionals designed the Class 11 Biology NCERT solutions, especially to help students get the marks they deserve. This leads to content that is tailored specifically to pupils, making it simple to get a thorough understanding of the subject.

Marks Weightage of Class 11 Biology NCERT

The final examination has a total of 70 marks. Unit 1 carries 7 points, Unit 2 carries 11 points, Unit 3 carries 15 points, Unit 4 carries 17 points, and Unit 5 carries 20 points. Since the fifth unit is worth the highest points, it is advisable for students to begin their preparations from this unit. Following unit 5, they can proceed to work on Unit 4. However, students should not neglect the other units and regularly revise them as well. 

Benefits of CBSE Class 11 Biology NCERT Solutions

Downloading the NCERT Solutions for Class 11 Biology  When students download the file containing Biology NCERT Class 11 solutions, they gain several benefits. Among these advantages are the following:

  • Students appearing for NEET should have a strong command of diagrams from the class 11th and 12th biology sections. The Class 11 Biology Solutions expose these diagrams with clarity and descriptions to help students understand complex diagrams 
  • As you are aware, the subject of class 11th is very essential; thus, a student should learn with great interest and develop the habit of asking questions in regular class; never keep doubts with you. NCERT Solutions For Class 11 Biology can be of assistance to you
  • Students would be  able to cover the class subject in less time than using printed notes or notes taken from another student; in fact, you could re-read and recall your whole chapter in a matter of minutes if you use your own handwritten notes using the NCERT Solutions as a guide
  • A major number of questions in JIPMER, AIIMS, and NEET are based on diagrams and the content of NCERT tables Therefore, bear this in mind
  • Use NCERT Solutions for Class 11 Biology to help you solve all of the NCERT textbook questions

How to utilise Class 11 Biology NCERT Solutions for Board and Competitive exams?

Did you know that roughly 85% of the questions in NEET 2019 are taken directly from the NCERT textbook? About 45 % of these question comes from the NCERT class 11 text book, while the remaining 65 % comes from the NCERT class 12 text book. This information clarifies the significance of the NCERT class 11 Biology text book. You will be prepared for any problem with NCERT Solutions for Class 12 Biology. Students can follow the tips given below and utilise the solutions to the maximum not just for their boards, but for future competitive examinations.

Note making: When attending a lecture, make a note of the important points that the teacher makes. This procedure stimulates your intellect and allows you to recall knowledge more easily. Furthermore, this is a crucial tool to employ when revising for your exam. Because you prepared these notes, you will be able to absorb and recall the information more easily than if you had used a textbook.

Information Chunking: It is impossible to remember every detail as it is owing to the vast curriculum. This is where a technique called chunking helps. The entire content is conveniently broken into chunks or smaller pieces of information. Not only does this make it easier to remember the points, but it also reduces cognitive load which reduces mental fatigue. You can split it according to your understanding.

Drawing: Diagrams play an important role in the Class 11 Biology NCERT curriculum. However, you will not have much time during the exam to sketch neatly or clearly, which may prevent you from receiving the marks you need to pass the exam. As a result, practise the essential schematics beforehand with various components such as shading and texturing. This will set you apart from the rest of your classmates. This may take some effort to learn and master, but the payoff is well worth it.

How would the extramarks study material help in learning more effectively?

  • At Extramarks, we have a staff of highly qualified experts who are specialists in their disciplines. All answers are worked on by these professional lecturers
  • All solutions in the NCERT Solutions for Class 11 Biology are totally accurate and stated in points. This guarantees that pupils may readily comprehend the content
  • These solutions are written in accordance with CBSE norms. This will assist the students to score well-deserving marks in their final examinations

Q.1 What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.

Ans.

Tidal volume is the volume of air that is inhaled or exhaled during normal respiration. It is about 500 mL. A healthy person breathes about 12-16 times in a minute.

The approximate value of tidal volume for a healthy human per hour can be calculated as follows:

The tidal volume of a healthy human in an hour = Tidal volume per breath X Number of times an adult human being breathes per minute X 60 minutes

= 500 mL X 12-16 times per minute X 60 minutes

= 3,60,000 – 4,80,000 mL

Q.2 What is the average cell cycle span for a mammalian cell?

Ans.

The average cell cycle span for a typical mammalian cell is about 24 hours. These cells divide once every 24 hours to form two daughter cells.

Q.3 Distinguish cytokinesis from karyokinesis.

Ans.

Cytokinesis Karyokinesis
Cytokinesis is the process of division of cytoplasmic contents during mitotic or meiotic division. Karyokinesis is the division of nuclear material during the process of cell division.
It is the last step of cell division and requires completion of karyokinesis step. It proceeds well before cytokinesis.
No well-demarcated stages are seen. The process of karyokinesis has well-demarcated stages.

Q.4 Describe the events taking place during Interphase.

Ans.

During the course of the cell cycle, the time span a living cell spent in between two consecutive M- phase (mitotic- phase) is called Interphase. This is the metabolically active phase during which the cell grows in size, duplicating its genome (DNA) and accumulating nutrients required for mitosis. It is also called the resting phase as no change in chromatin structure is observed under the microscope during this phase. On the basis of molecular and biochemical events during Interphase, it has been divided into three sub-phases.

Gap 1 phase (G1): This is the duration between mitosis and initiation of DNA replication where the cell is metabolically active, and grow without DNA replication. This phase is marked by the increase in cell size where the cell prepares for DNA replication

Synthesis phase (S): This phase begins with DNA synthesis or DNA replication. By the end of this phase the DNA content of the cell doubles but without increase in the chromosomal number. Thus if the initial chromosomal number of a cell was 2n and DNA content was 2C, at the end of S-phase the DNA content will become 4C while chromosomal number remains the same (2n).In S-phase initiation of centriole duplication also takes place in the cytoplasm.

Gap 2 phases (G 2): In G2 phase the cell continues to grow and synthesize the regulatory proteins and enzymes necessary for mitotic phase.

Q.5 What is G0 (quiescent phase) of cell cycle?

Ans.

G0 or quiescent phase of the cell cycle is the phase when the cell has come out of G1 phase to enter an inactive stage. The cells are metabolically active however do not undergo division unless required depending upon the environmental condition. It is also known as the resting phase. The adult myocardiocytes which never undergo division of cells which divide rarely to replace cells lost due to some injury or cell death remain in G0 or quiescent phase of the cell cycle.

Q.6 Why is mitosis called equational division?

Ans.

Mitosis is called equational division because the two daughter cells formed at the end of mitosis acquire the same number and kinds of chromosomes as the parent cell nucleus. There is a duplication of each chromosome before the onset of mitosis and is composed of two chromatids which get separated equally during the process of mitosis. The orderly distribution of chromosomes on the mitotic spindle helps in equal separation of chromatids of each chromosome. Also, since there is no crossing over during the process of mitosis, it results in the same kinds of the chromosome as the parent cell.

Q.7 Name the stage of cell cycle at which one of the following events occur:

  1. Chromosomes are moved to spindle equator.
  2. Centromere splits and chromatids separate.
  3. Pairing between homologous chromosomes takes place.
  4. Crossing over between homologous chromosomes takes place.

Ans.

  1. Metaphase
  2. Anaphase
  3. Zygotene of Meiosis I
  4. Pachytene of Meiosis I

Q.8 Describe the following:

  1. Synapsis
  2. Bivalent
  3. Chiasmata.

Draw a diagram to illustrate your answer.

Ans.

  1. Synapsis: It is the pairing up of duplicated homologous chromosomes (one from each parent) during the process of meiosis in such a way that the DNA of the non-sister chromatids is aligned. It takes place during the zygotene stage of prophase I of meiosis. Synapsis of homologous chromosomes is accompanied by the formation of a complex structure called synaptonemal complex.
  2. Bivalent: It is the complex formed by a pair of synapsed homologous chromosomes. It is a bivalent because there are two chromosomes in close association. It is also known as tetrad because bivalent contains four chromatids. Bivalents condense and become visibly distinct during the pachytene stage of prophase I of meiosis. Crossing over occurs between the non-sister chromatids of the bivalent. This results in non-identical sister chromatids of a chromosome.
  3. Chiasmata: It is an X-shaped structure of chromosomes formed during the diplotene stage of meiosis I. The synaptonemal complex dissolves resulting in separation of recombined homologous chromosomes of the bivalents except at the site of crossing over.

Q.9 How does cytokinesis in plant cells differ from that in animal cells?

Ans.

In animal cells, a furrow develops in the plasma membrane. This cleavage furrow deepens gradually and ultimately joins in the centre resulting in the division of the cell cytoplasm into two halves. However, the plant cells are surrounded by a rigid and inextensible cell wall due to which no cell furrow formation takes place. The process of cytokinesis is very different in plant cells as compared to that of animal cells. The major differences are:

  1. Cell plate formation: The cell wall formation begins with the precursor called cell plate at the centre of the cell.
  2. Extension of cell plate: The cell plate extends outwardly in all direction and ultimately unites with the exiting lateral wall and thus function as middle lamella between the walls of two adjacent cells
  3. Deposition of cell wall components: Once the cell plate is fused to the cell wall of the cell, deposition of cell wall components such as cellulose takes place to make a proper cell wall.

Q.10 Find examples where the four daughter cells differ from meiosis are equal in size and where they are found unequal in size?

Ans.

An example where equal-sized daughter cells are formed at the end of meiosis: Spermatogenesis results in the production of equal-sized haploid sperms.

An example where unequal sized daughter cells are formed at the end of meiosis: During the process of oogenesis, unequal sized daughter cells are formed at the end of meiosis.

Q.11 Distinguish anaphase of mitosis from anaphase I of meiosis.

Ans.

Anaphase of Mitosis Anaphase I of Meiosis
Centromere joining the sister chromatids splits and the two daughter chromatids (produced during the S phase), now referred to as chromosomes segregate to opposite poles. The homologous chromosome separate to the opposite poles while the sister chromatids remain attached as there is no splitting of the centromere.
This separation maintains the DNA content and the chromosome number in the daughter cells. This separation results in the reduction of chromosome number to half as the homologous chromosome separate to opposite poles.

Q.12 List the main differences between mitosis and meiosis.

Ans.

Mitosis Meiosis
Mitosis is a process of cell duplication during which one cell gives rise to two daughter cells. Meiosis is a division of a germ cell involving two divisions of the nucleus and giving rise to four gametes, or sex cells, each possessing half the number of chromosomes of the original cell.
The steps of mitosis are Interphase, Prophase, Metaphase, Anaphase, Telophase and Cytokinesis The steps of meiosis are Interphase, Prophase I, Metaphase I, Anaphase I, Telophase I, Prophase II, Metaphase II, Anaphase II and Telophase II.
Single division results in the formation of two diploid daughter cells. A series of two divisions result in the formation of four haploid daughter cells.
Daughter cells are genetically identical to the mother cell. Daughter cells are not genetically identical to the mother cell.
Since the number and kind of chromosomes remain the same as mother cells, it is also known as equational division. Since the number of chromosomes reduces to half at the end of meiosis I, it is known as reductional division.
The pairing of homologous chromosomes does not occur. The pairing of homologous chromosomes and cross-over of chromosomes occurs.
Crossing over does not occur. Crossing over between non sister chromatids occurs.
Mitosis takes place during cellular reproduction & general growth and repair of the body. Meiosis takes place during sexual reproduction.
It occurs in all kinds of cells except sex cells. It occurs only in sex cells resulting in the formation of gametes.
It occurs in all organisms. It is a characteristic feature of Humans, animals, plants, fungi.

Q.13 What is the significance of meiosis?

Ans.

Significance of meiosis:

  • Since gametes in sexually reproducing organisms are produced by the process of meiosis (resulting in the reduction of chromosome number to half), it helps in conserving the specific chromosome number of each species across generations.
  • It also increases genetic variability due to the process of crossing over happening during the process. This results in gaining new characteristics by the population over generations. Such variations help in the process of evolution.

Q.14 Can there be mitosis without DNA replication in ‘S’phase?

Ans.

No, mitosis cannot take place without DNA replication in ‘S’phase. Mitosis is an equational division resulting in the production of two genetically identical daughter cells. This is achieved because the DNA of each chromosome duplicates by the process of replication during the S or synthetic phase of the cell cycle. The amount of DNA per cell doubles. This is followed by segregation of the two chromatids of each chromosome during the anaphase stage of mitosis bringing the DNA content identical to that of the parent cell.

Q.15 Can there be DNA replication without cell division?

Ans.

Yes, there can be DNA replication without cell division. Endoreplication is the replication of DNA of the cell without the cell division. This results in a large number of copies of the same DNA molecule(s) within the cell nucleus resulting in polyploidy. One example is polytene chromosomes of the salivary glands of fruit flies. The repeated division of chromosomes without any cell division results in a large number of sister chromatids tightly bound to each other. DNA replication without cell division is a common phenomenon in plants.

Q.16 Analyze the events during every stage of cell cycle and notice how the following two parameters change

  1. Number of chromosomes (N) per cell
  2. Amount of DNA content (C) per cell

Ans.

Mitotic Cell Cycle:

  • The number of chromosomes (N) per cell: Number of chromosomes (N) per cell never changes in the mitosis. If the cell has diploid (2n) number of chromosomes, even after the synthetic phase where the DNA replication occurs, the chromosome number remains 2n.
  • Amount of DNA content (C) per cell: During the S phase (synthetic) of interphase in mitosis, DNA synthesis or replication takes place. It results in duplication of the amount of DNA per cell i.e. from the initial amount of 2C, it becomes 4C. It is during the anaphase that the sister chromatids separate due to splitting of the centromere, thereby bringing the DNA content per cell back to 2C amount.

Meiotic Cell Cycle:

  • The number of chromosomes (N) per cell: Meiosis results in the reduction of the number of chromosome per cell to half. The cell starts with a 2N number. Before meiosis, the DNA replication occurs and each chromosome gets 2 sister chromatids. It is during the anaphase I of meiosis I that homologous chromosomes segregate reducing the chromosome number to N. There is no further reduction in the chromosome number in meiosis II.
  • Amount of DNA content (C) per cell: Amount of DNA content per cell is also reduced to half during the process of meiosis. Let us assume the initial DNA content is to be 1C per cell before the onset of meiosis. During the replication process before the onset of meiosis I, each chromosome undergoes replication thereby doubling the DNA content (2C). During anaphase I, homologous chromosomes segregate to opposite poles however there is no splitting of the centromere. The DNA content of the 2 daughter cells produced becomes 1C again. It is during the anaphase II in meiosis that the centromeres split resulting in separation of sister chromatids further into 4 daughter cells. This results in a reduction of DNA content to half as compared to the parent cell.

Q.17 Define the following:
(a) Exocrine gland
(b) Endocrine gland
(c) Hormone

Ans.

(a) Exocrine gland: The glands which secrete their products via a duct are called exocrine glands. Examples: sweat glands, salivary glands, mammary glands and liver.

(b) Endocrine gland: The glands that do not secrete their products (hormones) via any duct, but directly secrete them into the bloodstream or into its surrounding tissues are known as endocrine glands. Examples: adrenal gland, pituitary gland and thyroid glands.

(c) Hormone: The secretions of endocrine glands that perform their function at a distant location, away from the location of their origin by travelling through the bloodstream are called hormones. They can also diffuse to their neighbouring tissues.

Current scientific definition defines hormones as “Hormones are non-nutrient chemicals which act as intercellular messengers and are produced in trace amounts.” Example: thyroid hormone, follicle stimulating hormone, testosterone.

Q.18 Diagrammatically indicate the location of the various endocrine glands in our body.

Ans.

Diagrammatic representation of the location of various endocrine glands in our body:

Q.19 List the hormones secreted by the following:
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(l) G-I Tract

Ans.

(a) Hypothalamus: Two types of hormones are secreted by hypothalamus:

(i) Releasing hormones: Gonadotropin releasing hormone, thyrotrophin releasing hormone, somatotrophin releasing hormone and adrenocorticotrophic hormone

(ii) Inhibiting hormones: Somatostatin, growth- inhibiting hormone and melanocyte-stimulating hormone

(b) Pituitary: List of hormones secreted by two different parts of the pituitary gland:

(i) Hormones secreted by Adenohypophysis: Growth hormone (GH), prolactin, thyroid stimulating hormone (TSH), adrenocorticotrophic hormone (ACTH), luteinizing hormone (LH), melanocyte stimulating hormone (MSH) and follicle stimulating hormone (FSH)

(ii) Hormones secreted by Neurohypophysis: Oxytocin and vasopressin

(c) Thyroid: Thyroxin (T4), triiodothyronine (T3) and thyrocalcitonin

(d) Parathyroid: Parathyroid hormone (PTH)

(e) Adrenal: List of hormones secreted by adrenal gland:

  • Corticoid hormones are secreted by adrenal cortex. Corticoids are further divided into

-glucocorticoid (hormone secreted is cortisol)

-mineralocorticoid (hormone secreted is aldosterone).

  • Catecholamines are secreted by adrenal medulla. It secretes adrenaline (also called epinephrine) and noradrenaline (also called norepinephrine).

(f) Pancreas: Insulin and glucagon

(g) Testis: Testosterone

(h) Ovary: Estrogen and progesterone

(i) Thymus: Thymosins

(j) Atrium: Atrial natriuretic hormone (ANH)

(k) Kidney: Erythropoietin

(l) G-I Tract: Gastrin, secretin, cholecystokinin and gastric inhibitory peptide (GIP, also known as glucose-dependent insulinotropic peptide).

Q.20 Fill in the blanks:


Ans.

Q.21 Write short notes on the functions of the following hormones:
(a) Parathyroid hormone (PTH)
(b) Thyroid hormones
(c) Thymosins
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon

Ans.

(a) Parathyroid hormone (PTH): Parathyroid hormone is a peptide hormone and secreted by parathyroid glands. It plays a significant role in:

  • Calcium homeostasis in the body. It increases the calcium level in blood. It also stimulates calcium absorption from digested food and reabsorption of Ca++ by renal tubules.
  • It stimulates the dissolution and demineralisation of bones.

(b) Thyroid hormones: Thyroxine (T4), tri-iodothyronine (T3) and thyrocalcitonin are collectively called as thyroid hormones. Iodine is essential for an adequate rate of thyroid hormone synthesis. Low level of thyroid hormones (hypothyroidism) results in enlargement of the thyroid gland, commonly called goitre. Excessive synthesis of thyroid hormones is called hyperthyroidism which is also harmful to the body. Thyroid hormones are essential for many physiological processes.

  • They play an important role in maintaining normal basal metabolic rate (BMR).
  • They control the metabolism of carbohydrates, proteins and fats.
  • They also help in red blood cell (RBCs) formation and regulating blood calcium levels.
  • They also influence water and electrolytes balance.

(c) Thymosins: It is a peptide hormone and secreted by the thymus. Its functions are:

  • Differentiation of T-lymphocytes which are essential for cell-mediated immunity.
  • Promotes antibody production which provides humoral immunity.

(d) Androgens: Androgens are a group of hormones. Main androgen hormone is testosterone hormone. Androgens are secreted by leydig cells of testes. The main functions of androgens are:

  • It stimulates the development of male secondary sex characteristics.
  • It is required for the development, maturation and functions of the male accessory sex organs like epididymis, vas deferens, seminal vesicles, prostate gland, urethra, etc.
  • It stimulates spermatogenesis (formation of spermatozoa) and influence male sexual behaviour (libido).

(e) Estrogens: Estrogen is a steroid hormone produced by the ovaries. The main functions of estrogen are:

  • It stimulates the development of female secondary sex organs.
  • It helps in the development of ovarian follicles and mammary glands.
  • It also regulates female sexual behaviour.

(f) Insulin and Glucagon: Functions of Insulin and glucagon:

  • Insulin stimulates uptake of excess glucose and its utilisation by liver cells by the process of glycogenesis. In glycogenesis, excess glucose is converted into its polymer glycogen and stored in the liver cells. Deficiency of insulin results in a disease, called diabetes mellitus.
  • Glucagon acts in just the opposite manner. In case of low glucose level in blood, it stimulates the breakdown of glycogen and release of glucose in the bloodstream. It also stimulates the synthesis of glucose from non-carbohydrate substrates like pyruvate, lactate, glycerol, etc., by the process of gluconeogenesis.

Q.22 Give example(s) of:
(a) Hyperglycemic hormone and hypoglycemic hormone
(b) Hypercalcemic hormone
(c) Gonadotrophic hormones
(d) Progestational hormone
(e) Blood pressure lowering hormone
(f) Androgens and estrogens

Ans.

(a) Hyperglycemic hormone and hypoglycemic hormone: Hyperglycemic hormone- Glucagon, Hypoglycemic hormone- Insulin

(b) Hypercalcemic hormone: Parathyroid hormone

(c) Gonadotrophic hormones: Luteinising hormone, follicle-stimulating hormone

(d) Progestational hormone: Progesterone

(e) Blood pressure lowering hormone: Atrial natriuretic factor (ANF)

(f) Androgens and estrogens: Androgen- Testosterone, Estrogen- Estradiol

Q.23 Which hormonal deficiency is responsible for the following?
(a) Diabetes mellitus
(b) Goitre
(c) Cretinism

Ans.

(a) Diabetes mellitus: Diabetes mellitus is caused by deficiency of insulin. It is also caused when insulin becomes non-responsive to blood glucose level.

(b) Goitre:Goitre is the enlargement of thyroid glands and happens due to the deficiency of thyroid hormones.

(c) Cretinism: It is caused by low level of thyroid during pregnancy that results in the stunted growth of baby.

Q.24 Briefly mention the mechanism of action of FSH.

Ans.

Follicle Stimulating Hormone (FSH) is synthesised and secreted by the anterior lobe of the pituitary gland. FSH, being a glycoprotein, is insoluble in lipid thus, cannot enter the target cells. So, these glycoproteins first bind with the specific receptor molecules, that are found on the surface of a cell membrane, form a hormone-receptor complex and activate the cellular system to perform functions.

Mechanism of Action: After the release of FSH from the site of production, it reaches the ovary, uterus or testis and binds to FSH receptor, present at the plasma membrane of these organs. FHS receptor is a transmembrane G-protein coupled receptor protein. The binding of FSH to receptor generates second messengers (cAMP). Secondary messengers start several biochemical changes inside these organs and carry out many important functions.

Functions:

  • In the ovary, it results in follicular development.
  • In the uterus, it is involved in the development of secretory endometrium during the luteal phase.
  • In the males, the FSH is responsible for spermatogenesis.

Q.25 Match the following:

Column I Column II
(a) T4 (i) Hypothalamus
(b) PTH (ii) Thyroid
(c) GnRH (iii) Pituitary
(d) LH (iv) Parathyroid

Ans.

Column I Column II
(a) T4 (ii) Thyroid
(b) PTH (iv) Parathyroid
(c) GnRH (i) Hypothalamus
(d) LH (iii) Pituitary

Q.26 Choose the correct answer among the following:

(a) Gastric juice contains

(i) pepsin, lipase and rennin

(ii) trypsin, lipase and rennin

(iii) trypsin, pepsin and lipase

(iv) trypsin, pepsin and renin

(b) Succus entericus is the name given to

(i) a junction between ileum and large intestine

(ii) intestinal juice

(iii) swelling in the gut

(iv) appendix

Ans.

(a) (i) Gastric juice contains pepsin, lipase and rennin

(b) (ii) Succus entericus is the name given to intestinal juice

Q.27 Match column I with column II


Ans.

Q.28 Answer briefly:

(a) Why are villi present in the intestine and not in the stomach?

(b) How does pepsinogen change into its active form?

(c) What are the basic layers of the wall of alimentary canal?

(d) How does bile help in the digestion of fats?

Ans.

(a) The main function of villi is to increase the surface area of intestine for increased and efficient absorption of digested nutrients (monosaccharide and amino acids) through diffusion. Absorption of nutrients through diffusion is only possible after the complete digestion of food. Complete digestion of food occurs only in the small intestine, not in the stomach. That’s why villi are present in the intestine and not in the stomach.

(b) Hydrochloric acid (HCl), secreted by parietal or oxyntic cells of the gastric gland in stomach, converts pepsinogen into its active form pepsin.

(c) The alimentary canal wall consists of four basic layers namely:

  • Serosa: It is made up of a thin mesothelium and some connective tissues.
  • Muscularis: It consists of smooth muscle cells and is arranged into inner circular and outer longitudinal layers. Circular and longitudinal layers are responsible for mixing and propelling of food respectively along the gastrointestinal tract.
  • Sub-mucosal layer: It consists of loose connective tissues which contain nerves, blood and lymph vessels. Glands are also found in duodenal mucosa.
  • Mucosal layer: It forms the innermost layer. It is irregular with folds (rugae) in stomach and villi in the small intestine. Villi increase the surface area for the absorption of digested nutrients. Villi are further covered with microscopic projection called microvilli which gives a brush border appearance to the epithelial lining of villi.

(d) Fats are insoluble in water and aggregate to form a large mass. Lipases cannot act on these aggregates. Bile is amphipathic in nature with both hydrophilic and hydrophobic groups. Bile breaks these large fat globules into small droplets by the process of emulsification and makes them water-soluble. Hydrophobic parts of bile interact with fat and hydrophilic region interacts with water molecules that make these fat droplets suspended in an aqueous environment. Now, these water-soluble fat droplets are easily accessible to lipases that hydrolyse them into monoglycerides.

Q.29 State the role of pancreatic juice in digestion of proteins.

Ans.

Pancreatic juice contains several pro-enzymes required for the digestion of proteins – trypsinogen, chymotrypsinogen and procarboxypeptidases. First trypsinogen is converted into active trypsin by an enzyme called enterokinase present in intestinal mucosa. Trypsin further activates other protein digestive enzymes.

Trypsin, Chymotrypsin and carboxypeptidases further act on proteins, proteases and peptones and convert it into dipeptides.

Q.30 Describe the process of digestion of protein in stomach.

Ans.

The stomach stores up to 2 litres of partially digested food. Food remains in the stomach for 4-5 hours. The muscular walls of the stomach contract vigorously and mix food with juices that are secreted whenever food enters the stomach. Gastric juice secreted by glands of the stomach contains HCl and inactive proteolytic enzyme- pepsinogen. HCl activates the conversion of pepsinogen into active pepsin which hydrolyses proteins into smaller peptides (proteoses and peptones). HCl also provides an acidic environment (pH of about 2.0) required for optimal activity of pepsin. Rennin is another proteolytic enzyme found only in infants that helps in the digestion of milk by hydrolysing milk protein, casein. At the end of this process, the contents of the stomach acquire a thick, soupy consistency called chyme. The chyme goes into the first part of the small intestine whenever the sphincter, a narrow opening at the base of the stomach, relaxes.

Q.31 Bile juice contains no digestive enzymes, yet it is important for digestion. Why?

Ans.

Bile juice is digestive secretion released by the gall bladder. It helps in the digestion of fat. It does not contain any digestive enzyme, but contains amphipathic bile salts- bilirubin and biliverdin that are essential for solubilisation of large fat globules by breaking them into smaller droplets which can mix with water. This process is called emulsification. Emulsified fat is further hydrolysed by lipases which are otherwise not able to act on large fat globules. Bile juice is alkaline, so it also activates lipases.

Q.32 Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?

Ans.

Chymotrypsin is synthesised in pancreas and secreted in the small intestine with other proteolytic enzymes in the form of pancreatic juice. Chymotrypsin is synthesized in an inactive form as chymotrypsinogen which is cleaved by trypsin to give active chymotrypsin. Chymotrypsin helps in digestion of proteins. It hydrolyses proteins into dipeptides. Other digestive enzymes secreted by same source glands are trypsinogen and procarboxypeptidase which also help in digestion of proteins.

Q.33 How are polysaccharides and disaccharides digested?

Ans.

Digestion of polysaccharides (carbohydrates) starts in the mouth and completes in the small intestine. Enzymes required for the digestion of polysaccharides are called amylases. Approximately 30% of the polysaccharides are digested in the mouth by salivary amylases which are secreted into the mouth by salivary glands. Salivary amylases work at pH 6.8; however when they reach stomach, they become inactive due to the low pH of the stomach. Digestion of polysaccharides again starts in the small intestine where pancreatic amylase breaks down large polysaccharides into small disaccharides.

Disaccharides are further digested into monosaccharides by other enzymes present in pancreatic juice.

Q.34 What would happen if HCl were not secreted in the stomach?

Ans.

HCl is secreted by parietal cells of the stomach which maintains the acidic pH of the stomach. HCl is required for the activation of pepsin which is required for the digestion of proteins. Pepsin is secreted in the stomach as inactive pepsinogen. HCl triggers the conversion of pepsinogen into pepsin. Further, pepsin needs low pH (1.5 to 2.0) for its optimal activity which is provided by HCl. Pepsin becomes inactive at pH above 6.5. Therefore, if HCl is not secreted by the stomach, pepsin will be activated and this will affect protein digestion.

Q.35 How does butter in your food get digested and absorbed in the body?

Ans.

Butter mainly consists of fat (fat 80%, water 15% and protein 5%). Butter is water-insoluble and forms large fat globules in the aqueous environment of stomach.

Digestion: When these large fat globules reach small intestine, they are first converted into small fat droplets by the action of bile juice. This process is called emulsification. Bile juice also activates pancreatic lipases which are present in the small intestine. Emulsified fat droplets are hydrolysed into triglycerides by pancreatic lipases in the small intestine. These triglycerides are further hydrolyzed into diglycerides and monoglycerides by intestinal lipases.

Absorption: Fats are absorbed by membrane of enterocytes, present in intestinal mucosa, by simple diffusion as they are hydrophobic and can cross the lipid bilayer of cell membrane. Cytoplasm of enterocytes and blood are aqueous in nature and these monoglycerides are water immiscible. So these monoglycerides are packaged into water-soluble particles, called chylomicrons. Chylomicrons are lipo-proteins in nature. Chylomicrons are exocytosed by enterocytes. Exocytosed chylomicrons enter lymph vessel (lacteal). Lymph vessel transports chylomicrons to blood stream.

Q.36 Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.

Ans.

Digestion of proteins starts in the stomach and completes in the small intestine. As proteins travel along the gastro-intestinal tract, they are hydrolyzed into smaller peptides and finally into amino acids. Enzymes which hydrolyze proteins are called proteases. There are several proteases which are secreted by the glands of mucosa of stomach and small intestine. Most of these proteases are secreted in an inactive form and converted into active enzyme in the gastro-intestinal lumen.

Stomach: Stomach secretes gastric juice which contains proteolytic enzymes- pepsinogen and rennin along with HCl. HCl makes the gastric environment acidic which is required for the conversion of inactive pepsinogen into active pepsin. Pepsin hydrolyses proteins into smaller peptides. Stomach also secretes another protease rennin that helps in digestion of milk proteins in infants.

Small Intestine: As the proteins enter the small intestine, they are acted upon by the proteases present in pancreatic juice. Pancreatic juice contains several inactive enzymes required for the digestion of proteins –trypsinogen, chymotrypsinogen and procarboxypeptidases. First trypsinogen is converted into active trypsin by an enzyme, enterokinase present in intestinal mucosa. Trypsin further activates other protein digestive enzymes.

Q.37 Explain the term thecodont and diphyodont.

Ans.

Thecodont is a condition in which teeth are embedded in the sockets. These sockets are also called alveoli of the jawbones. These teeth have cylindrical roots. Mammals and crocodiles have thecodont teeth.

Diphyodont is a condition in which teeth appears in two sets in the entire life. The first set of teeth are called deciduous or temporary teeth. The deciduous teeth are replaced by the second and the final set of permanent teeth. For example, humans are diphyodont. They have 20 deciduous teeth which are replaced by 32 permanent teeth.

Q.38 Name different types of teeth and their number in an adult human.

Ans.

There are 32 teeth in an adult human. They are of four different types and this type of dentition is called heterodont dentition. These four types of teeth are Incisors, canines, premolars and molars.

Incisors: Incisors are front teeth. There are eight incisors, four in the upper jaw and four in the lower jaw.

Canines: Canines are situated on both sides of the incisors. There are 4 canines, two in the upper jaw and two in the lower jaw.

Premolars: Premolars are situated next to canines. They are eight in number, four in the upper jaw and four in the lower jaw.

Molars: Molars are situated next to premolars. They are twelve in numbers, six in the upper jaw and six in the lower jaw.

Q.39 What are the functions of liver?

Ans.

The functions of the liver are as follows:

  • Liver cells secrete bile juice which helps in the digestion of fat. Fat forms large insoluble fat globules in the aqueous environment of the stomach which are inaccessible to pancreatic lipases. Bile juice contains amphipathic bile salts (bilirubin and biliverdin) which break down large fat globule into small fat droplets. This process is called emulsification. These fat droplets are hydrolyzed into glycerides by pancreatic lipases.
  • Bile juice secreted from the liver is also required for activation of pancreatic lipases.
  • It detoxifies the blood by removing and metabolizing poisonous substances.
  • It produces heparin protein which prevents blood clotting inside the blood vessels.
  • It stores glucose as glycogen and breaks down glycogen to glucose between meals to maintain constant blood glucose levels by the process of glycogenesis.
  • It produces urea from amino groups and ammonia.
  • Liver is the site of vitamin-A synthesis from the carotenes of carrot.

Q.40 Define Glomerular Filtration Rate (GFR).

Ans.

Glomerular Filtration Rate (GFR) is the amount of filtrate formed by the kidneys per minute. It is measured in ml/min. In a healthy person, GFR is approximately 125 ml/min that is equal to 180 litres per day.

Q.41 Explain the autoregulatory mechanism of GFR.

Ans.

Kidneys maintain a uniform Glomerular Filtration Rate (GFR) with the help of the autoregulatory mechanism. Kidneys have a microscopic structure called juxta glomerular apparatus (JGA) located between an afferent arteriole and distal convoluted tube of the same nephron that is required for this auto-regulation. Whenever there is a fall in glomerular blood flow/glomerular blood pressure/GFR, cells of juxta glomerular apparatus release a hormone called renin which converts angiotensinogen in the blood to angiotensin I and further to angiotensin II.

  • Angiotensin II is a strong vasoconstrictor. It increases the glomerular blood pressure and blood flow and thereby, increases the GFR.
  • Angiotensin II also activates the release of Aldosterone which causes reabsorption of Na+ and water from the distal parts of the tubule. This also causes an increase in blood pressure and GFR.

Q.42 Indicate whether the following statements are true or false:
(a) Micturition is carried out by a reflex.
(b) ADH helps in water elimination, making the urine hypotonic.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.

Ans.

(a) True

(b) False

(c) True

(d) True

(e) True

Q.43 Give a brief account of the counter current mechanism.

Ans.

The transport of substances facilitated by a special arrangement of Henle’s loop of nephron and vasa recta is called counter current mechanism.

When fluid flows in these two tubes which are anti-parallel to each other, it forms a counter current. This helps mammals to produce concentrated urine. There are two counter current systems in human kidneys-

  • Henle’s loop consists of ascending and descending limbs. Fluid flows in the two limbs in opposite directions making a counter current system.
  • Second counter current system is made by vasa recta. Blood flows in the two limbs of vasa recta in anti-parallel direction.

Henle’s loop and vasa recta are positioned next to each other. Descending limb of Henle’s loop is anti-parallel to ascending limb of vasa recta and ascending limb of Henle’s loop is anti-parallel to descending limb of vasa recta. Counter current system and proximity of the loops help to increase the osmolarity (solute concentration) in inner medullary interstitium i.e. from 300mOsmolL-1 in the cortex to about 1200mOsmolL-1 in the inner medulla. Increased solute concentration in medullary interstitium causes water to come out of collecting duct making the urine concentrated. This system has the ability to concentrate urine upto four times depending upon the need.

Q.44 Describe the role of liver, lungs and skin in excretion.

Ans.

In addition to kidneys, liver, lungs and skin also play an important role in excretion of waste materials from the body.

Liver: It is the largest gland of the human body. Liver converts toxic ammonia into less toxic urea; thus carry out the detoxification role. It helps in the excretion of bile containing substances like bilirubin, biliverdin which are produced by the decomposition of haemoglobin pigment. Then bilirubin, biliverdin, urea along with other waste like cholesterol, degraded steroid hormones, vitamins and drugs are excreted by the liver. Finally, these substances are thrown out of the body along with digestive wastes.

Role of Lungs: Lungs diffuse out a large amount of gaseous waste like CO2 (18 litres/day) as part of normal respiration along with the significant quantity of water.

Role of Skin: Skin contains sweat and sebaceous glands which help in the excretion of many substances. The sweat produced by sweat glands secrete NaCl, small amounts of urea, lactic acid, etc., along with water. Sebaceous glands help in excretion by elimination of certain sterols, hydrocarbons and waxes through sebum.

Q.45 Explain micturition.

Ans.

Micturition is the process of releasing of urine from urinary bladder. Urine formed by kidney is carried to urinary bladder where it is stored. When urinary bladder is filled with urine, it exerts the pressure on bladder wall. Pressure activates receptors present in bladder wall that send signals to central nervous system (CNS). The CNS sends the signal for the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine.

An adult human excretes about 1 to 1.5 litres of urine per day. The urine is light yellow coloured watery fluid which is slightly acidic with a characteristic odour.

Q.46 Match the items of column I with those of column II:


Ans.

Q.47 What is meant by the term osmoregulation?

Ans.

Osmoregulation is a process by which organisms maintain homeostasis of water content in their body. It protects body fluids from becoming too dilute or too concentrated. For example, large amount of water goes into the nephrons from blood during the process of glomerular filtration. This water is reabsorbed by medullary interstitium and transferred again to blood in vasa recta thus, maintaining water homeostasis in blood.

Q.48 Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic. Why?

Ans.

Terrestrial animals are generally either ureotelic or uricotelic because it is an adaptation to conserve water. In these animals, water soluble ammonia is converted into less soluble urea and uric acid. These are then filtered by the kidney and excreted out with urine. This requires minimum water; thus the animal conserves water.

Q.49 What is the significance of juxta glomerular apparatus (JGA) in kidney function?

Ans.

Kidneys have a microscopic structure called juxta glomerular apparatus (JGA) between an afferent arteriole and distal convoluted tubule of the same nephron that is required for the maintenance of uniform Glomerular Filtration Rate (GFR). Whenever there is a fall in GFR, cells of juxta glomerular apparatus release a hormone called renin which travels into the bloodstream of glomerulus where it converts angiotensinogen into angiotensin I. Angiotensin I further gets converted to angiotensin II. Angiotensin helps in constriction of smooth muscle cells of the blood vessel that increases the blood pressure. Increased blood pressure results in an increase in glomerular filtration rate. Angiotensin also stimulates the secretion of aldosterone from adrenal glands. Aldosterone increases the absorption of sodium ions and water which increases blood pressure and brings the GFR back to normal. Thus, juxta glomerular apparatus plays a complex regulatory role in the functioning of the kidney.

Q.50 Name the following:
(a) A chordate animal having flame cells as excretory structures
(b) Cortical portions projecting between the medullary pyramids in the human kidney
(c) A loop of capillary running parallel to Henle’s loop

Ans.

(a) Amphioxus
(b) Columns of Bertini
(c) Vasa recta

Q.51 Fill in the gaps:
(a) Ascending limb of Henle’s loop is _______ to water whereas the descending limb is _______ to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone _______.
(c) Dialysis fluid contains all the constituents as in plasma except _______.
(d) A healthy adult human excretes (on an average) _______ gm of urea/day.

Ans.

(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.

(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone antidiuretic hormone or vasopressin.

(c) Dialysis fluid contains all the constituents as in plasma except the nitrogenous waste.

(d) A healthy adult human excretes (on an average) 25-30 gm of urea/day.

Q.52 Draw the diagram of a sarcomere of skeletal muscle showing different regions.

Ans.

The diagram of a sarcomere of skeletal muscle showing different regions:

Q.53 Define sliding filament theory of muscle contraction.

Ans.

According to the sliding filament theory of muscle contraction, contraction of muscle fibres to produce contractile force takes place by the sliding of the thin filaments (actin fibres) over the thick filaments (myosin) found in the sarcomere of the muscle cell.

[Explanation: Sliding filament theory best describes the molecular basis of muscle contraction. It explains how myofibrils (actin and myosin) interact to produce contractile force. In sarcomere, actin forms thin filaments and myosin forms thick filaments. Sarcomere consists of two bands- A-band (Dark band) and I-band (Light band). A-band is the region of myosin thick filaments while I-band is the region of actin filaments which are not superimposed with myosin filaments. Actin filaments of I-band are connected to an elastic fibre called Z-line. There is a region in the central part of sarcomere that does not overlap with actin fibres is called H-zone. During the process of muscle contraction, thick filaments (myosin fibres, A-band) remains constant while thin filaments (actin fibres, I-band) change their length and slide over the myosin filaments. Actin filaments interact with the myosin head that results in pulling the actin filaments towards the centre of the sarcomere. In this process, length of sarcomere decreases and Z-lines of sarcomere come closer.]

Q.54 Describe the important steps in muscle contraction.

Ans.

Muscle contraction is explained by sliding filament theory. According to sliding filament theory, contraction of muscle fibers to produce contractile force takes place by the sliding of the thin filament (actin fibers) over the thick filaments (myosin) found in the sarcomere of the muscle cell. Important steps in muscle contraction are as follows:

1. Central nervous system sends signal for initiation of muscle contraction via a motor neuron.

2. Neural signal reaches motor end plate (the junction between a motor neuron and sarcolemma of the muscle fiber) and releases a neurotransmitter (acetylcholine) which generates an action potential in the sarcolemma that spreads through the muscle fiber and causes the release of calcium ions into the sarcoplasm.

3. Calcium ions bind to troponin that removes tropomyosin from the active sites of actin. These exposed active actin sites are now available to interact with myosin.

5. Heads of myosins form cross bridges by interacting with active sites on actin filaments and pull them towards the center of A-band by utilising the energy from ATP hydrolysis. Z line which is attached to actin filaments are also pulled inwards that shortens the sarcomere resulting in contraction. In this process, I-bands get shortened while the lengths of A-bands remain the same.

6. Myosin goes into relaxed state by releasing ADP and Pi. Cross bridge is broken. A new ATP binds to myosin and upon ATP hydrolysis next cycle of cross-bridge formation starts. This process continues till the calcium ions are sent back into sarcoplasmic cisternae that results into masking of active sites on actin filaments.

Q.55 Write true or false. If false, change the statement so that it is true.
(a) Actin is present in thin filament
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.

Ans.

(a) True

(b) False. H-zone of striated muscle fibre represents only thick filaments.

(c) True

(d) False. There are 12 pairs of ribs in man.

(e) True

Q.56 Write the difference between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle

Ans.

  1. Actin and Myosin

  1. Red and White muscles
  1. Pectoral and Pelvic girdle

Q.57 Match Column I with Column II:

Column I Column II
(a) Smooth muscle (i) Myoglobin
(b) Tropomyosin (ii) Thin filament
(c) Red muscle (iii) Sutures
(d) Skull (iv) Involuntary

Ans.

Column I Column II
(a) Smooth muscle (iv) Involuntary
(b) Tropomyosin (ii) Thin filament
(c) Red muscle (i) Myoglobin
(d) Skull (iii) Sutures

Q.58 What are the different types of movements exhibited by the cells of human body?

Ans.

Four different types of movements are exhibited by the cells of human body:

1. Amoeboid movement: This kind of movement is shown by specialised cells like macrophages and leukocytes in the blood which migrate from bloodstream to the site of injury to perform their function. These cells move by the formation of pseudopodia (streaming of cytoplasm) in the direction of movement. Microfilaments of the cytoskeleton are involved in this kind of movement.

2. Ciliary movement: Ciliary movement is shown by cells of the ciliated epithelium lining internal tubular organs. Cells present in the lining of trachea perform ciliary movement to sweep the dirt and mucus out of the lung. In females, the ciliary movement shown by the cells lining the fallopian tubes helps in moving ova from ovary to uterus. Cilia are also found in the cochlear cells of the ear.

3. Muscular movement: Muscle cells exhibit contraction and relaxation movement. This kind of movement is involved in moving our limbs, jaws, tongue, etc.

4. Flagellar movement: Sperms swim by means of flagellum which shows flagellar movement.

Q.59 How do you distinguish between a skeletal muscle and a cardiac muscle?

Ans.

Q.60 Name the type of joint between the following:-
(a) atlas/axis
(b) carpal/metacarpal of thumb
(c) between phalanges
(d) femur/acetabulum
(e) between cranial bones
(f) between pubic bones in the pelvic girdle

Ans.

Bone Type of Joint
(a) atlas/axis Pivotal Joint
(b) carpal/metacarpal of thumb Saddle Joint
(c) between phalanges Hinge Joint
(d) femur/acetabulum Ball and socket Joint
(e) between cranial bones Fibrous joint
(f) between pubic bones in the pelvic girdle Cartilaginous joint (Pubic Symphysis)

Q.61 Fill in the blank spaces:
(a) All mammals (except a few) have __________ cervical vertebra.
(b) The number of phalanges in each limb of human is __________.
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely __________ and __________.
(d) In a muscle fibre Ca++ is stored in __________.
(e) __________ and __________ pairs of ribs are called floating ribs.
(f) The human cranium is made of __________ bones.

Ans.

(a) All mammals (except a few) have seven cervical vertebra.

(b) The number of phalanges in each limb of human is 14.

(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely troponin and tropomyosin.

(d) In a muscle fibre Ca++ is stored in sarcoplasmic reticulum.

(e) 11th and 12th pairs of ribs are called floating ribs.

(f) The human cranium is made of eight bones.

Q.62 ‘All elements that are present in a plant need not be essential to its survival’. Comment.

Ans.

The statement that “all the elements that are present in a plant need not be essential to its survival” holds true as only the elements which fulfill the below-mentioned criteria are essential for the survival of the plant, rest are not essential.

  • The element must be absolutely necessary for supporting growth and reproduction. In the absence of these elements, plants cannot grow or produce seeds.
  • The element cannot be replaced with any other element which means that the requirement is very specific. Deficiency of that particular element cannot be met by replacing it with other elements.
  • The element must be directly involved in the metabolism of the plant.

There are cases where elements like gold, selenium, etc., have been found in plants but these are not essential.

Q.63 Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?

Ans.

The purification of water and nutrient salts is very important in studies involving mineral nutrients using hydroponics because the presence of impurities in a mineral solution may alter the concentration of the solution therefore, altering the optimum growth conditions like pH required for Hydroponics. Due to such alterations, the standardised mineral solution might be rendered non-useful for the growth of the plant. These impurities could be in the form of inorganic elements or compounds that are detrimental to the plants. For example, Sodium (Na) if greater than 70 ppm, Boron (B) if greater than 2.0 ppm, anions of chloride (Cl) if greater than 100 ppm and bicarbonate (HCO3) if greater than 40 ppm can be very harmful to the growth of the plants. Also, in order to determine the exact nature and amount of mineral required, it is necessary that the nutrient is present in known amounts. Any kind of impurity from water or nutrient salt will hamper the validity of the experiment.

Q.64 Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.

Ans.

Macronutrients: They have the following features:

  • Essential for the growth of the plant
  • Required in large amounts
  • Directly involved in the metabolism of the plant
  • Cannot be replaced with any other nutrients
  • Generally present in large amounts (in excess of 10 mmole Kg-1 of dry matter).

Example: carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur etc.

Micronutrients: They have the following features:

  • Essential for the growth of the plant
  • Required in small amounts
  • Directly involved in the metabolism of the plant
  • Cannot be replaced with any other nutrient
  • Generally present in very small amounts (less than 10 mmole Kg-1 of dry matter).

Example: iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

Beneficial Nutrients: The elements which are not essential but their presence enhances the growth of the plants are known as beneficial nutrients. In addition to the essential elements, beneficial elements are also required by some higher plants for better growth. These elements have been reported to enhance the resistance of the plant to biotic and abiotic stresses. Example: sodium, silicon, cobalt and selenium.

Essential Elements: Essential elements are elements that have the following features:

  • Absolutely necessary for supporting normal plant growth
  • Cannot be replaced by any other element
  • Directly involved in the metabolism of the plant
  • Can be macronutrients or micronutrients (depending upon the quantities in which they are required by the plant)
  • 17 essential elements which carry out essential functions in the plants
  • Classified into four subgroups based on the function.

Example: carbon, hydrogen, oxygen, nitrogen, iron, manganese, copper, molybdenum, zinc.

Q.65 Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.

Ans.

The different deficiency symptoms in plants are as follows:

  • Chlorosis is the loss of chlorophyll leading to yellowing of leaves. This is caused by the deficiency of elements like N, K, Mg, S, Fe, Mn, Zn and Mo.
  • Necrosis is the death of tissues, particularly leaf tissue. This is caused by the deficiency of elements like Ca, Mg, Cu, K.
  • Delayed flowering is the result of low levels of elements like N, S or Mo.
  • Stunted plant growth is the result of deficiencies of Cu and K.
  • Inhibition of cell division is caused by lack or low levels of N, K, S or Mo.

[N-Nitrogen, K-Potassium, Mg-Magnesium, S-Sulphur, Fe-Iron, Mn-Manganese, Ca-Calcium, Cu-Copper, Zn-Zinc and Mo-Molybdenum]

Q.66 If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?

Ans.

If a plant shows a symptom which could develop due to deficiency of more than one nutrient, the real deficient mineral element can be found by following steps:

  • Observe the morphological changes in all the parts of the plant.
  • Compare the changes with the available standard tables.
  • Design an experiment where one can add the nutrient one by one and then look for the recovery of the plants after addition of the particular element.

This strategy helps in finding out the real deficient mineral element.

Q.67 Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?

Ans.

The reason behind the appearance of deficiency symptoms first in younger parts of the plant as compared to that of mature organs is due to the difference in the mobility of the element under study. When the elements are relatively immobile and are not transported out of the mature organs, the deficiency symptoms first appear in younger growing parts. For example, elements like sulphur and calcium are part of the structural component of the cell and are not easily released from the older tissues making it deficient in the younger and newer organs. The elements that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms first appear in the older tissues. For example, the deficiency symptoms of nitrogen, potassium and magnesium are visible first in the senescent leaves. In the older leaves, biomolecules containing these elements are broken down, making it available for younger leaves.

Q.68 How are the minerals absorbed by the plants?

Ans.

The main organ of the plant through which minerals are absorbed from the soil is root. The process of absorption has two well-demarcated phases.

  • In the initial phase, rapid uptake of ions occurs passively into the space outside the cell membrane of the cells (known as apoplast). This uptake occurs through ion-channels which are transmembrane proteins that function as selective pores.
  • In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ or the space inside the plasma membrane (known as symplast) of the cell. This process is energy-driven and thus an active process. The movement of ions is called flux.

Q.69 What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N2 -fixation?

Ans.

Nitrogen is one of the most prevalent elements in living organisms. The atmospheric nitrogen cannot be absorbed by plants as such – it needs to be fixed as nitrates before plants can absorb and transport it to leaves. Reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation and few prokaryotic species are capable of fixing nitrogen. These prokaryotic microbes, which have the enzyme nitrogenase, are capable of nitrogen reduction and are called N2 fixers. Rhizobium is one such N2 fixer. The conditions necessary for fixation of atmospheric nitrogen by Rhizobium are as follows:

  1. The most important condition required for fixation of atmospheric nitrogen by Rhizobium is the presence of the anaerobic condition. The enzyme nitrogenase, that is required for the conversion of atmospheric nitrogen to ammonia cannot function in the presence of oxygen. The nodule where the bacteria reside, provide the anaerobic condition.
  2. The presence of leguminous haemoglobin, an oxygen scavenger is required for the process of nitrogen fixation to occur.
  3. The process of ammonia synthesis by nitrogenase requires a high input of energy and this is obtained from the respiration of the host cells.

Role in N2-fixation: The free-living aerobic bacteria Rhizobium enters into a symbiotic relationship with leguminous plants and starts fixing free nitrogen of the atmosphere into ammonia which can be utilised by the host plant. The important features of Rhizobium are

  • It occurs inside the anaerobic nodule which carries all the necessary biochemical components such as the enzyme nitrogenase and leghaemoglobin, which is an oxygen scavenger.
  • The enzyme nitrogenase of Rhizobium is a Mo-Fe protein and catalyzes the below-written reaction. Ammonia is the first stable product of nitrogen fixation. The reaction is as follows-

N2+ 8e+ 8H+ + 16ATP → 2NH3+ H2+ 16ADP+16Pi

  • The ATP required for the above reaction is provided by the host.
  • Ammonia formed following nitrogen fixation is incorporated into amino acids as the amino groups.

Q.70 What are the steps involved in formation of a root nodule?

Ans.

The formation of nodule requires a series of interaction between the Rhizobium, which is free-living aerobic bacteria and the roots of the host leguminous plants. Rhizobium is capable of fixing free nitrogen of the atmosphere into ammonia which can be utilised by the host plant. This reaction occurs inside the anaerobic conditions maintained inside nodule which carries all the necessary biochemical components. Main stages in the formation of the nodule are described below:

  • Rhizobia multiply, colonise the root surroundings and get attached to epidermal and root hair cells.
  • Root hairs curl and bacteria invade the root hair.
  • Infection thread carries the bacteria into the root cortex, where nodule formation is initiated.
  • Bacteria are released from the infection thread into the cells – specialised nitrogen-fixing cells start differentiating.
  • The nodule is formed – it has a direct vascular connection with the host for exchange of nutrients.

Q.71 Which of the following statements are true? If false, correct them:

(a) Boron deficiency leads to stout axis.

(b) Every mineral element that is present in a cell is needed by the cell.

(c) Nitrogen as a nutrient element, is highly immobile in the plants.

(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.

Ans.

(a) True

(b) False. All the mineral elements that are present in a cell are not needed by the cell. For example, gold and selenium are found accumulated in plants but they are not used in any plant process.

(c) False. Nitrogen as a nutrient is highly mobile in the plants. Its deficiency results in deficiency symptoms to show up first in older tissues.

(d) False. It is very difficult to establish the essentiality of micronutrients because of the trace requirements

Q.72 What is meant by modification of root? What type of modification of root is found in the:

(a) Banyan tree
(b) Turnip
(c) Mangrove trees

Ans.

The root performs two major functions in a plant:

  1. Absorption of nutrients and water from the soil for proper growth of the plants, and
  2. Providing mechanical support to plant by anchoring it tightly with the soil

However, in some cases, depending on the environmental conditions and physiological needs of the plants, the roots get modified to perform the functions apart from the ones mentioned above. This change in the root structure and function is known as root modification. The types of root modification in different plants/trees are listed below:

S.No. Plants/trees with root modification Purpose of modification
(a) Banyan tree Support: In a banyan tree, the prop roots arise from aerial branches and provide support to the plant.
(b) Turnip Storage: In turnip, the roots get swollen and function as a storage organ.
(c) Mangrove tree Respiration: In mangrove trees which grow in marshy areas, the roots get modified to pneumatophores that help in gaseous exchange.

Q.73 Justify the following statements on the basis of external features:

(i) Underground parts of a plant are not always roots.

(ii) Flower is a modified shoot.

Ans.

(i) Underground parts of a plant are not always roots.

During the course of evolution, plants modified their external morphologies according to the environmental conditions of their ecological niches for better adaptability. There are several examples where the stem, which is generally the aerial part of a plant, gets modified in a way that it remains under the soil. Such stem modifications can easily be recognized by the presence of nodes and internodes on them. The main reason for such modifications is to protect the plant from unfavourable environmental conditions. For example, the underground part of ginger, turmeric, zaminkand, and Colocasia all have distinct node and internodes, thus they are stem modifications. Similarly, in banana, pineapple and Chrysanthemum, the stem remains underground and grows beneath the soil. In case of onion and garlic, the leaves remain underground, become fleshy, and store food. Therefore, from the above examples, it is clear that underground parts of plants are not always roots as both stem and leaves in some cases have modified themselves to grow beneath the ground.

(ii) The flower is a modified shoot.

During the flowering season, the apical shoot meristem gets transformed to floral meristem. The close anatomical study of a flower reveals the presence of nodes and internodes. The main axis gets highly condensed and the internodes do not undergo further elongation. From the apex, instead of leaves, various floral appendages such as calyx, corolla, androecium and gynoecium arise. Thus, it can be said that a flower is a modified shoot.

Q.74 How is a pinnately compound leaf different from a palmately compound leaf?

Ans.

Pinnately compound leaf Palmately compound leaf
It is a compound leaf where leaflets are arranged like bird’s feathers on a common axis, called the rachis.

Example: Neem

In this type of leaf, the leaflets are attached at a common point, that is, they radiate out from the tip of petiole.

Example: Silk cotton

Q.75 Explain with suitable examples the different types of phyllotaxy.

Ans.

The term phyllotaxy originated from the Greek words ‘phyllon’ which stands for “leaf” and “taxis” which stands for arrangement. Thus, the pattern of arrangement or organization of leaves on a branch or stem is called phyllotaxy.

Broadly, it is of three types:

(i) Alternate: When each node of the axis bears a single leaf, it is called alternate phyllotaxy. Example: China rose, mustard and sunflower plants.

(ii) Opposite: When each node of the axis bears a pair of leaf facing in the opposite direction, it is called opposite phyllotaxy. Example: Calotropis and guava plants.

(iii) Whorled: When more than two leaves arise from each node, it is called whorled phyllotaxy. Example: Alstonia.

Q.76 Define the following terms:
(a) aestivation
(b) placentation
(c) actinomorphic
(d) zygomorphic
(e) superior ovary
(f) perigynous flower
(g) epipetalous stamen

Ans.

(a) Aestivation: The positional arrangement of sepals or petals in the floral bud with respect to other sepals or petals in the same whorl is called aestivation. The four different types of aestivation are twisted, valvate, imbricate and vexillary.

(b) Placentation: The arrangement of ovules in the ovary is called placentation. Depending on the pattern of arrangement, placentation is categorized into different types, namely marginal, parietal, axile, basal, central and free central.

(c) Actinomorphic: The flowers that can be divided into equal halves in any of the radial plane passing through the centre of the flower are called actinomorphic. Thus, these flowers show radial symmetry. Example: flowers of chilli and Datura are actinomorphic.

(d) Zygomorphic: The flowers that can be divided into two equal halves only in one particular vertical plane are called zygomorphic. These flowers show bilateral symmetry. Example: flowers of Gulmohar, bean and Cassia are zygomorphic.

(e) Superior ovary: The flowers in which the gynoecium is present at the highest position, that is, above calyx, corolla and androecium, the ovary in such flowers is called the superior ovary. Example: mustard, china rose and brinjal.

(f) Perigynous flower: The flower in which the ovary (gynoecium) is located at the centre of the thalamus while other floral appendages are located at the rim of the thalamus, such flower is called perigynous flower. Example: plum, rose, and peach.

(g) Epipetalous stamen: When the stamen is connected to the petals of a flower, it is known as the epipetalous stamen. Example: Brinjal

Q.77 Differentiate between

(a) Racemose and cymose inflorescence

(b) Fibrous root and adventitious root

(c) Apocarpous and syncarpous ovary

Ans.

(a) Racemose and cymose inflorescence

Racemose inflorescence Cymose inflorescence
1. The main floral axis shows unlimited growth and the terminal flower is not formed. The main axis has limited growth and terminates in a flower.
2. The inflorescence follows an acropetal order, where the flowers start maturing from the bottom of fluorescence axis. The inflorescence follows basipetal succession and the flower at the top of axis matures first.

(b) Fibrous root and adventitious root

Fibrous root Adventitious root
1. When the roots arise from the base of stem by replacing the primary root, it is called fibrous root. When the roots arise from any part of the plant other than radicle, it is called adventitious root. They may arise from stem or leaves.
2. These are generally short and very dense and are usually found in monocot plants. Example: maize. These may be very large and strong. Example: prop root of the banyan tree.

(c) Apocarpous and syncarpous ovary

Apocarpous ovary Syncarpous ovary
1. An ovary of a flower that has many free carpels terminating in a single ovary is known as the apocarpous ovary. When a gynoecium has many carpels that are fused into a single structure, it is called the syncarpous ovary.
2. Example: Lotus and rose. Example: Mustard and tomato.

Q.78 Draw the labelled diagram of the following:

(i) gram seed

(ii) V.S. of maize seed

Ans.

(i) Gram seed

(ii) Vertical section (V.S.) of maize seed

Q.79 Describe modifications of stem with suitable examples.

Ans.

The stem is the aerial part of a plant which originates from the plumule of the embryo and bear branches, leaves, flowers and fruits. In certain cases, it gets modified into tendrils, thorns, fleshy, or flattened structures. The main functions of stem are as follows:

(i) It forms an axis to support the lateral branches, flower and fruits.

(ii) It conducts water and nutrients through vascular tissue.

(iii) It bears leaves.

Apart from the above-mentioned functions, the stems sometimes get structurally modified to perform the following specialized functions:

(i) Storage: In the case of potato, zaminkand, Colocasia, ginger, and turmeric the stem gets modified to form an underground structure that protects the plant from unfavourable conditions and also acts as a storage organ.

(ii) Support: In some creepers such as cucumber, pumpkin and watermelon, the stem gets modified to form a slender and spirally coiled structure, called tendril, that coils around the wall or rope and helps the plant in climbing up.

(iii) Protection: In Bougainvillea and Citrus plants, the axillary bud of stem gets modified to a straight and pointed woody structure called a thorn, which protects the plants from being eaten up by herbivorous animals.

(iv) Photosynthesis: In arid regions, to prevent high water loss through transpiration, plants modify their stem into a flattened or fleshy cylindrical structure. These structures have chlorophyll and perform photosynthesis. Examples: Opuntia and Euphorbia.

(v) Vegetative propagation: In plants like grasses and strawberry, the underground stem spreads to new niche and develops into new plants. In mint and jasmine, the lateral branches after growing for certain length, arch down to the ground and form a new plant. In some aquatic plants like Pistia and Eichhornia, each node bears the rosette of leaves and tuff of roots, which give rise to a new plant once it gets detached from the parent plant.

Q.80 Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagram after studying them.

Ans.

Fabaceae Solanaceae
Name of Plant Pea plant (Pisum sativum) Makoi plant (Solanum nigrum)
Vegetative Characters Herbaceous climber Annual herb
Root Contains root nodules Tap root without nodules
Stem Erect, branched, solid, possesses tendrils Herbaceous, erect, smooth and branched
Leaves Alternate, pinnately compound, reticulate venation, pulvinate base, terminal leaflet modifies into tendrils Alternate, simple, estipulate with reticulate venation
Inflorescence Racemose Cymose
Flower Bisexual, zygomorphic Bisexual, actinomorphic
Calyx Five sepals, gamosepalous; imbricate aestivation Five sepals, united, persistent; valvate aestivation
Corolla Five petals, polypetalous, vexillary aestivation Five petals, united, valvate aestivation
Androecium Ten stamens, diadelphous, anther dithecous Five epipetalous stamens
Gynoecium Superior ovary, mono carpellary, unilocular with many ovules, single style Bicarpellary, syncarpous, superior ovary, bilocular, swollen placenta with many ovules.
Fruits Legume Berry
Seeds Several, non-endospermic Many, endospermic
Floral diagram

Q.81 Describe the various types of placentations found in flowering plants.

Ans.

The arrangement of ovules in the ovary is called placentation. Depending on the mode of arrangement, placentation is of following types:

  1. Marginal placentation: It is observed in the flowers where gynoecium is monocarpellary, apocarpous and unilocular as in pea. At the ventral surface of the ovary, a ridge is formed by the placenta and ovules are borne in the ridge.
  2. Axile placentation: It is observed when gynoecium is multicarpellary, syncarpous and multilocular. Here, the placenta is axial and ovules are attached to it in the multilocular ovary. Example: China rose, tomato and lemon.
  3. Parietal placentation: It is observed when gynoecium is multicarpellary, syncarpous and unilocular. Here, ovules develop on the periphery of the ovary. Example: mustard and Argemone.
  4. Free central placentation: In this, the gynoecium is multicarpellary, syncarpous and unilocular and the ovules are borne on the central axis without any septa. Example: Dianthus and Primrose.
  5. Basal placentation: In this, the gynoecium is unilocular with placenta developing at the base of the ovary and single ovule attaches to it. Example: Sunflower and marigold.

Q.82 What is a flower? Describe the parts of a typical angiosperm flower.

Ans.

The reproductive unit of the angiosperm plant, bearing male or female or both reproductive parts that are responsible for sexual reproduction is called flower. A flower is a modified stem with a highly condensed axis.

A typical angiosperm flower consists of four whorls, namely calyx, corolla, androecium and gynoecium. The flower, having all these four whorls, is known as a complete flower. The complete flower is a bisexual flower as both male and female gamete forming organs are present in it. The various parts of a typical angiosperm flower are described below:

(i) Calyx: It is the outermost whorl of the flower. It consists of individual units called sepals. Generally, sepals are green in colour. Its main function is to protect the bud of the plant from adverse environmental conditions. If the sepals are free, it is called polysepalous but sometimes sepals are fused giving rise to the gamosepalous condition.

(ii) Corolla: Next to calyx is corolla, which consists of individual petals. Petals are beautiful coloured structures that play a key role in attracting insects to increase the chances of pollination.

(iii) Androecium: The male gamete producing part of the plant is called androecium. Its individual units are called the stamen. An individual stamen consists of filament and anther.

(iv) Gynoecium: It is the female reproductive organ of a plant. It consists of carpels. Each carpel is made up of three parts, namely; ovary, style and stigma.

Q.83 How do the various leaf modifications help plants?

Ans.

Leaves are the photosynthetic sites of the plant; however, in some cases, they get modified to provide support, protection, storage or nutrients to the plants.

  1. Support: In case of plants like pea, the leaves get converted to a spiral structure called tendrils that help the plant to climb up on the support structure like a wall.
  2. Protection: In cactus, the leaves get reduced to spins to protect the plant from animals as well as to reduce water loss through stomata.
  3. Storage: The fleshy underground leaves of garlic and onion help in the storage of food and ensure the survival of plant under unfavourable conditions.
  4. Nutrients: In the carnivorous plants such as venus-fly trap that grows in the nitrogen-deficient soil, the leaves get modified into a pitcher to capture the insect and thus provide required nutrients to the plant.

Q.84 Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.

Ans.

The arrangement of flowers on the floral axis is called inflorescence.

During the flowering season of a fully developed plant, the apical meristem gives rise to floral meristem which in turn develops the floral axis. Floral axis contains laterally developed flowers. On the basis of the type of growth of the floral axis, the inflorescence is of two types:

(i) Cymose inflorescence or determinate inflorescence: When the main axis terminates in a flower, it is known as cymose inflorescence. This type of inflorescence limits the growth.

(ii) Racemose or indeterminate inflorescence: When the main axis keeps on growing without terminating in a flower bud, it is known as racemose inflorescence.

Q.85 Write the floral formula of a actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carpels with superior ovary and axile placentation.

Ans.

Keys:

Terms Symbol used in the floral formula
Actinomorphic
Bisexual
Five united sepals K(5)
Five free petals C5
Five free stamens A5
Two united carpels with superior ovary and axile placentation G(2)

Q.86 Describe the arrangement of floral members in relation to their insertion on thalamus.

Ans.

The floral appendages are arranged on the thalamus. On the basis of the location of the ovary with respect to other floral appendages (calyx, corolla and androecium), the flower can be broadly classified into three categories:

(a) Hypogynous flower: In some plants like mustard, china rose and brinjal, gynoecium occupies the highest position compared to calyx, corolla and androecium. Such flowers are called hypogynous flowers. Here, the ovary is located at the uppermost part of the thalamus. Such an ovary is called the superior ovary.

(b) Perigynous flower: In the flowers like rose, the gynoecium is at the centre of the thalamus and other floral whorls are at the rim of the thalamus with ovary situated at the same level. Such flowers are called perigynous flowers.

(c) Epigynous flower: In the flowers like guava, the margin of thalamus completely covers the ovary and other floral whorls arise above the ovary. Such flowers are called epigynous flowers.

Q.87 By looking at a plant externally can you tell whether a plant is C3 or C4? Why and how?

Ans.

No, one cannot tell whether a plant is C3 or C4 by looking at only external features. There is no morphological difference between leaves of C3 and C4 plants, so it is almost impossible to differentiate these plants just by external investigation. C4 plant can be differentiated from C3 plant by anatomical examination only. The vascular bundle of C4 plants are surrounded by large cells called bundle sheath cells. Such structure found especially in C4 plants is called ‘Kranz Anatomy’. Thus, the study of vertical section of leaves under compound microscope is necessary to differentiate C3 plants from C4 plants.

Q.88 By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.

Ans.

The major anatomical difference between C3 and C4 plant is the Kranz anatomy (Kranz is a German word for “wreath”). The leaves of C4 plant show the presence of Kranz anatomy, which is absent in C3 plants. Generally, the cross-sections of C3 leaves under a microscope shows only one type of cells that contains chloroplast. These are loosely arranged and are called mesophyll cells. But in case of C4 plants, apart from being loosely arranged mesophyll cells, a tightly arranged and chloroplast containing cells are found surrounding the vascular bundle. These cells are called bundle sheath cells. This anatomical feature of C4 plant is called Kranz anatomy.

Q.89 Even though a very few cells in a C4 plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?

Ans.

Calvin cycle occurs in all photosynthetic plants and the main enzyme involved is RuBP carboxylase-oxygenase (RuBisCO). It is during the Calvin cycle that the sugar is synthesised by fixing carbon dioxide. The rate at which carbon dioxide is fixed determines the productivity of the plant. In case of C3 plants, all the mesophyll cells contain the enzyme RuBisCO, while in C4 plants only bundle sheath cell contain RuBisCO, thus it can be said that in C4 plant only a few cells carry out Calvin cycle and fix carbon dioxide as compared to C3 plants. Still, productivity is higher in C4 plants. The reason is the difference in the anatomy of these two plants.

RuBisCO binds to both carbon dioxide and oxygen but it has a higher affinity for carbon dioxide. Under normal condition, it binds to carbon dioxide and carries out the Calvin cycle for carbon fixation. But at high concentration of oxygen, RuBisCo catalyses the binding of oxygen to RuBP forming phosphoglycolate and initiates photorespiratory pathway. This pathway results in the release of carbon dioxide consuming ATP. There is no net synthesis of ATP or sugar molecule in the photorespiratory pathway. Photorespiration occurs only in C3 plants.

In C4 plant, the Kranz anatomy indirectly helps in the suspension of photorespiratory pathways in two ways-

  1. It lacks RuBisCO in its mesophyll cells thus minimising the chances of photorespiration pathway.
  2. The C4 acid formed in mesophyll cell is transported to bundle sheath cells, where they are decarboxylated to C3 acid and carbon dioxide molecule. The C3 acid is again transported back to mesophyll cells and help in the formation of more C4 acid. The carbon dioxide released in bundle sheath cell increases its concentration and thus minimizing the chances to oxygenase activity of RuBisCO enzyme.

Thus the productivity of C4 plant is higher as compared to C3 plants as they bypass the photorespiration pathway and fix more carbon under similar condition.

Q.90 RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C4 plants?

Ans.

RuBisCo is an enzyme that acts both as a carboxylase and oxygenase. However, it carries out more carboxylation in C4 plants due to following two reasons:

  1. C4 plants contain RuBisCO in bundle sheath cells but lack the same in mesophyll cells. This minimises the chances of an increase in oxygen concentration near the enzyme, in turn inhibiting its oxygenation activity.
  2. The C4 acid formed in mesophyll cell is transported to bundle sheath cells, where it is decarboxylated to C3 acid and carbon dioxide molecule. The C3 acid is again transported back to mesophyll cells and helps in the formation of more C4 acid. The carbon dioxide released in the bundle sheath cell increases its concentration and thus accelerates the carboxylation activity of the enzyme.

Q.91 Suppose there were plants that had a high concentration of Chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?

Ans.

The plant having a high concentration of chlorophyll b, but lacking chlorophyll a will be unable to carry out photosynthesis efficiently. The chlorophyll a molecules form the reaction centers of both Photosystem I and Photosystem II. The excited reaction center chlorophyll molecule transfers its energy to various acceptors in cyclic and non-cyclic photophosphorylation and thus NADPH2 molecule is produced. A plant lacking chlorophyll a molecule will be defective in light reaction of photosynthesis and in turn will not be able to carry out fixation of carbon dioxide.

Plants contain chlorophyll b and other accessory pigments for performing the following two functions:

  • They form the antenna molecule, absorb light energy and transfer it to reaction center chlorophyll a molecule, thus increasing the efficacy of the photosystem.
  • They surround the reaction center chlorophyll a molecule and thus protect it from photo-oxidative damage.

Q.92 Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?

Ans.

Photosynthesis is a light-dependent process and in the presence of light, plants synthesise the photosynthetic molecules like chlorophyll a and b. Due to their short life and also due to photo-oxidative damage of chlorophyll by light, plants need a continuous synthesis of chlorophyll molecules. When a plant is shifted from light to dark, it stops synthesising the chlorophyll molecule and the existing chlorophyll molecule loses its stability and gets degraded.

Under these conditions, the colour of accessory pigments like xanthophylls (yellow) and carotenoids (yellow to yellow-orange) predominates which do not require light for their synthesis. Thus, pigments carotenoids and xanthophyll are more stable and the leaves turn yellow or pale green under dark.

Q.93 Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?

Ans.

The rate of photosynthesis depends upon the external environmental factors like water supply, temperature, the concentration of carbon dioxide and light intensity as well as internal factors like chlorophyll and concentration of RuBisCO. To increase the photosynthetic efficiency plants can modulate internal factors, but they cannot modulate external factors.

When a plant is shifted from shade to light, it senses an increase in light intensity, which stimulate the plant to increase the rate of photosynthesis by increasing the chlorophyll content in leaves.

Thus, the leaves of the same plant on the shady side as compared to the sunny side of the potted plant kept in sunlight possess higher chlorophyll, thereby appearing darker green in colour.

Q.94 Figure 13.10 shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions:

(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?

Ans.

(a) Keeping all the other factors apart and considering light as the only factor affecting the rate of photosynthesis, in the above curve at point “A”, light is the limiting factor because the photosynthesis rate is minimum at this point.

(b) Apart from light, other environmental factors such as the concentration of carbon dioxide, water supply and temperature may also be the limiting factors. Also, the internal factors such as chlorophyll content in leaves may be one of the limiting factors affecting the rate of photosynthesis at the point “A”.

(c) Stage ‘C’ is representing that beyond it light is not a limiting factor.

At stage ‘D’, the rate of photosynthesis becomes saturated. This suggests that above it, the intensity of light does not play a role in increasing the rate of photosynthsis.

Q.95 Give a comparision between the following:

(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4 plants

Ans.

(a) C3 and C4 pathways

(b) Cyclic and non-cyclic photophosphorylation

(c) Anatomy of leaf in C3 and C4 plants

Q.96 What is the basis of classification of algae?

Ans.

The algal classification into three different groups is based on the following characteristics:

  • Nature of pigmentation
  • Type of photosynthetic food reserve
  • Flagella type
  • Cell wall structure and composition

[Note: The algae are divided into three main classes: Chlorophyceae, Phaeophyceae and Rhodophyceae.]

Q.97 When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?

Ans.

Reduction division is a type of cell division in which the chromosome number is reduced to half in the process of forming gametes/ spores for sexual reproduction. It is also called as meiosis.

Liverwort: In liverworts, the main plant-body is haploid (gametophytic). It bears the male and female sex organs which produce gametes. These gametes fuse to form a zygote. The zygote develops on the gametophytic plant-body to form a sporophyte. The sporophyte is differentiated into the foot, seta, and capsule. Reduction division takes place inside the capsule to form haploid spores.

Moss: In mosses, the gametophytic primary protonema develops into the secondary protonema which bears the sex organs which produce gametes. These gametes fuse to form a zygote. The zygote develops into a sporophyte. Reduction division in the capsule leads to haploid spore formation.

Fern: In ferns, the main plant-body is sporophytic. Its leaves are known as sporophylls and these bear the sporangia. Reduction division takes place in these sporangia, thereby producing many spores.

Gymnosperm: In gymnosperms, the main plant-body is sporophytic. They bear two types of leaves – microsporophylls and megasporophylls which produce the microsporangia and megasporangia respectively. Reduction division takes place in the microsporangia and megasporangia to produce the microspores/ pollen grains and megaspores/ eggs respectively.

Angiosperm: In angiosperms, the main plant-body is sporophytic and bears flowers. The male sex organ in the flower is the stamen, while the female sex organ is the pistil and both are present in the flower. Reduction division takes place in the anthers of the stamen (producing haploid pollen grains) and in the ovary of the pistil (producing haploid eggs).

Q.98 Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.

Ans.

Archegonia is a female sex organ that produces female gametes. The three groups that show distinct archegonia are:

  • Bryophytes,
  • Pteridophytes and
  • Gymnosperms

Life cycle of a liverwort, a bryophyte:

The life of a liverwort starts from the germination of a haploid spore to produce a protonema, which is either a mass of thread-like filaments or else a flattened thallus. The protonema is a transitory stage in the life of a liverwort, from which will grow the mature gametophore plant that produces the sex organs. The male organs are known as antheridia and produce sperm cells. The female organs are known as archegonia. Each archegonium has a slender hollow tube down which the sperm swims to reach the egg cell.

When sperm reach the archegonia, fertilisation occurs, leading to the production of a diploid sporophyte. After fertilisation, the immature sporophyte within the archegonium develops three distinct regions a foot, capsule and a seta. When the sporophyte has developed the archegonium ruptures to release the haploid spores by meiosis/ reduction division. The haploid spores will germinate and the life cycle will start again.

Q.99 Mention the ploidy of the following: protonemal cell of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; meristem cell of monocot, ovum of a liverwort, and zygote of a fern.

Ans.

Description Ploidy
Protonemal cell of a moss Haploid
Primary endosperm nucleus in dicot Triploid
Leaf cell of a moss Haploid
Prothallus cell of a fern Haploid
Gemma cell in Marchantia Haploid
Meristem cell of monocot Diploid
Ovum of a liverwort Haploid
Zygote of a fern Diploid

Q.100 Write a note on the economic importance of algae and gymnosperms.

Ans.

Economic importance of Algae:

  1. Algae as food – Algae species are used as food and are found to be rich in proteins, vitamins (A, B, C and E), lipids and minerals. Laminaria species is an edible seaweed in Japan.
  2. Algae as fodder for cattle – Algae is also used as food for cattle for example spirulina.
  3. Algae as fertilisers – Blue-green algae are treated as bio-fertilisers. Nostoc, Oscillatoria, Scytonema, Spirulina, etc. are used as fertilisers in rice fields. All these algae help to fix atmospheric Nitrogen in the ground.
  4. Agar-Agar: Agar-agar is a jelly-like substance of great economic value. It is obtained from certain red algae. Agar is used as a culture medium for growing callus in tissue culture as well as in confectionaries.
  5. Antibiotics and Medicines: Antibiotic Chlorellin, obtained from Chlorella is effective against several pathogenic bacteria. Seaweeds have a beneficial effect on gall bladders, pancreas, kidneys, uterus and thyroid glands.

Economic importance of gymnosperms:

  1. Gymnosperms like pine, fir, spruce, and cedar are conifers that are used for lumber. The wood is used in packaging industries and also for making home furnishing.
  2. Ornamental use of gymnosperms like for making bonsai and for Christmas decorations.
  3. Resins of coniferous trees are used to make turpentine oil, wood methanol, disinfectants, paints perfumes etc.

Q.101 Both gymnosperms and angiosperms bear seeds, then why are they classified separately?

Ans.

Both gymnosperms and angiosperms bear seeds but they are classified separately due to the following differences between the two groups:

Angiosperm Gymnosperm
Sporophylls form flowers. Sporophylls aggregate to form compact cones.
The microsporophylls consist of filaments called stamens and anthers contain pollen grains. The microsporophylls are not distinguished into filaments and anthers.
The megasporophylls are delicate consisting of ovary, style and stigma. The ovary contains the eggs. The megasporophylls are woody and do not contain the ovary, style and stigma. Therefore the eggs lie exposed.
Archegonia is replaced by an egg apparatus. Archegonia is present.
Two male gametes enter the egg apparatus at the time of fertilisation. One forms the zygote and the other forms the triploid endosperm. Single fertilisation event takes place to form the zygote. The endosperm is haploid.
The seeds develop inside a fruit. The seeds produced are naked as there is no fruit formation.
Examples: Apple, Guava, Peach Examples: Pine, Cedar, Fir, Ginkgo

Q.102 What is heterospory? Briefly comment on its significance. Give two examples?

Ans.

Heterospory is a phenomenon wherein two types of spores are produced by a plant as seen in pteridophytes. The two types of spores – megaspore and microspore – germinate into female and male gametophytes, respectively. The male gametophyte fuses with the female gametophyte which is retained on the parent sporophyte. The development of the embryos takes place in the female gametophyte.

This has evolutionary significance. This type of fertilisation and development could be the precursor to the phenomenon of seed development as seen in higher plants like angiosperms.

Examples: Genera like Selaginella and Salvinia are heterosporous and produce macro and microspores.

Q.103 Explain briefly the following terms with suitable examples:-

(i) protonema

(ii) antheridium

(iii) archegonium

(iv) diplontic

(v) sporophyll

(vi) isogamy

Ans.

(i) Protonema: Protonema is the first stage in the life cycle of moss. The protonema develops directly from a spore. The protonema is a mass of thread-like filaments or a flattened thallus. The protonema is a transitory stage in the life of a liverwort, from which will grow the mature gametophore plant that produces the sex organs.

(ii) Antheridium: The male sex organs in bryophytes are called antheridium. They produce biflagellate antherozoids which swim up the archegonium to fertilize the female gametes to form a zygote.

(iii) Archegonium: The female organ known as archegonia is flask-shaped and produces a single egg. The antherozoids (male gametes) are released into the water where they come in contact with archegonium. Each archegonium has a slender hollow tube in which the sperm swims to reach the egg cell. An antherozoid fuses with the egg to produce the zygote.

(iv) Diplontic: A diplontic organism is one in which the somatic cells are diploid (having two sets of chromosomes) while its gametes are haploid (having one set of chromosomes). The diploid part is the dominant, photosynthetic, independent phase of the plant. This kind of lifecycle is termed as diplontic. All seed-bearing plants i.e. gymnosperms and angiosperms, follow this pattern.

(v) Sporophyll: Sporophyll is a leaf that bears spores/ sporangia. In heterosporous plants, the sporophylls can bear either microsporangia (megasporophylls) or megasporangia (microsporophylls). Examples of plants that have sporophylls are ferns and some types of algae. In some cases, sporophylls may form distinct compact structures called strobili or cones (eg. Selaginella, Equisetum).

(vi) Isogamy: Isogamy is a form of sexual reproduction that involves gametes of similar morphology (similar shape and size), differing only in allele expression in one or more mating-type regions. Because both gametes look alike, they cannot be classified as “male” or “female.” Instead, organisms undergoing isogamy are said to have different mating types, most commonly noted as “+” and “-” strains, for example Chlamydomonas and other plants.

Q.104 Differentiate between the following:-

(i) red algae and brown algae

(ii) liverworts and moss

(iii)homosporous and heterosporouspteridophyte

(iv) syngamy and triple fusion

Ans.

(i)

Red Algae Brown Algae
1 Belong to class Rhodophyceae Belong to class Phaeophyceae
2 Contain Chlorophyll a, d and r-phycoerythrin Contain Chlorophyll a, c and fucoxanthin
3 Food is stored as floridean starch Food is stored as mannitol and laminarin
4 Flagella absent Flagella present
5 Cell wall made of cellulose, pectin and phycocolloids Cell wall made of cellulose and algin
6 Produce hydrocolloids called carrageen Produce hydrocolloids called algin

(ii)

Liverworts Moss
1 Leaves do not have veins Leaves have veins
2 Leaves have lobes and cilia arranged in two rows Leaves are serrated and spirally arranged
3 Sporophyte less elaborate Sporophyte more elaborate
4 Gemmae present Gemmae absent

(iii)

Homosporous Pteridophyte Heterosporous Pteridophyte
1 They bear spores that are of the same type. They bear two kinds of spores called megaspores and microspores.
2 The spores give rise to bisexual gametophytes. The megaspores and microspores give rise to female and male gametophyte respectively.

(iv)

Syngamy Triple Fusion
1 The fusion of the male gamete with the female gamete to form the zygote in angiosperms. The fusion of the male gamete with diploid secondary nucleus to form the triploid primary endosperm nucleus (PEN) in angiosperms.
2 The zygote develops into an embryo (with one or two cotyledons) The PEN develops into endosperm which provides nourishment to the developing embryo.

Q.105 How would you distinguish monocots from dicots?

Ans.

Monocots and dicots are the two classes of angiosperms. They can be distinguished by both anatomical and morphological characteristics. Some of the key features that can help distinguish the two are stated below:

Morphological Characteristic Monocot Dicot
Number of cotyledons in the seed One Two
Major leaf veins Parallel Reticulated
Roots Fibrous root Tap root
Flowers Grow in multiples of 3 Grow in multiples of 4 or 5
Anatomical Characteristics Monocot Dicot
Stem vascular bundles Scattered Bundled
Secondary growth Secondary growth absent Secondary growth often present

Q.106 Match the followings (column I with column II)

Column I Column II
(a) Chlamydomonas (i) Moss
(b) Cycas (ii) Pteridophyte
(c) Selaginella (iii) Algae
(d) Sphagnum (iv) Gymnosperm

Ans.

Column I Column II
(a) Chlamydomonas (iii) Algae
(b) Cycas (iv) Gymnosperm
(c) Selaginella (ii) Pteridophyte
(d) Sphagnum (i) Moss

Q.107 Describe the important characteristics of gymnosperms.

Ans.

Gymnosperms mean naked seeds (gymnos – naked; sperma – seed). Some important characteristics of gymnosperms are listed below:

  1. The seeds of these plants are naked as in they are not enclosed in fruits.
  2. The plants can vary in size ranging from a small shrub to the largest trees known on earth – Sequoia (height of 286 feet (87 m) or more, a circumference of 113 feet (34 m) or more).
  3. They have taproots and many plants have roots with the fungal association in the form of mycorrhiza. Some plant roots are also associated with N2 fixing bacteria.
  4. The leaves of gymnosperms, for example, pine trees, can withstand harsh weather and extreme cold due to reduced surface area, thick cuticle and sunken stomata.
  5. Gymnosperms are heterosporous; they produce haploid microspores and megaspores.
  6. The two types of spores, male and female are arranged on separate cones (also called strobili) to form the male and female cones. Fertilization occurs by wind pollination and results in the formation of uncovered seeds in the female cones.

Q.108 Differentiate between

(a) Respiration and Combustion

(b) Glycolysis and Krebs’ cycle

(c) Aerobic respiration and Fermentation F

Ans.

(a)

Respiration Combustion
1. It occurs within living cells only. It does not occur inside a living system.
2. It requires enzymes. It does not require enzymes.
3. It occurs in a highly regulated mode under controlled condition. It is non-regulated and uncontrolled.
4. It produces energy equivalents in the form of high energy ATP molecules. It produces energy in terms of heat and light only.

(b)

Glycolysis Krebs’ Cycle
1. It occurs in the cytoplasm. It occurs in the mitochondria.
2. It is a non-cyclic process. It is a cyclic process.
3. It is common in both aerobic and anaerobic respiration. It takes place only in aerobic respiration.
4. Less productive in terms of ATP and NADP generation. Produces 8 ATP molecules from one glucose molecule Produces 15 ATP molecules from one molecule of Acetyl CoA

(c)

Aerobic respiration Fermentation
1. It occurs in the presence of molecular oxygen only. It does not require molecular oxygen.
2. It takes place in both cytoplasm and mitochondria. It takes place only in the cytoplasm.
3. It is highly efficient and produces 38 ATP molecules per molecule of glucose. It is non-economical, produces only 2 molecules of ATP per molecule of glucose
4. The final products formed from one glucose molecule are carbon dioxide and water. The final products formed from one glucose molecule are ethyl alcohol and carbon dioxide

Q.109 What are respiratory substrates? Name the most common respiratory substrate.

Ans.

The complex organic compound that gets oxidized in the cell during respiration to release large amounts of energy is called respiratory substrate. Under normal condition, glucose is the most common respiratory substrate which is a carbohydrate along with six carbon atoms.

Q.110 Give the schematic representation of glycolysis?

Ans.

Schematic representation of glycolysis:

Q.111 What are the main steps in aerobic respiration? Where does it take place?

Ans.

The four main steps of aerobic respiration are as follows:

S.No Steps of aerobic respiration Site of occurrence in the cell
1. Glycolysis Cytoplasm
2. Krebs’ Cycle Mitochondrial matrix
3. Electron transport chain Inner membrane of mitochondria
4. Oxidative phosphorylation F0-F1 particle of cristae present in the inner membrane of mitochondria

Q.112 Give the schematic representation of an overall view of Krebs’ cycle.

Ans.

Schematic representation of Krebs’ cycle:

Q.113 Explain ETS.

Ans.

The electrons removed from the substrates of glycolysis and the Krebs’ cycle are stored in the reduction equivalents, namely NADH2 and FADH2. This energy is released when NADH2 and FADH2 are oxidized by passing their electrons to a chain of electrons carrier complex called Electron transport system, present in the inner membrane of mitochondria. These complexes transfer the electron through a series of redox reactions with high energy electrons entering the system and low-energy electrons leaving the system. The energy released through this process is utilized to pump out protons which develop a proton gradient (Proton motive force) across the inner membrane. This proton motive force is utilized by ATP synthase to generate high energy ATP molecules at 3 different sites.

Process: The NADH2 produced during the citric acid cycle are oxidized by an NADH dehydrogenase (complex I), and electrons are then transferred to ubiquinone which gets reduced. Ubiquinone also receives reducing equivalents via FADH2 (complex II). The reduced ubiquinone (ubiquinol) is then re-oxidized by transferring its electrons to cytochrome c via cytochrome bc1 complex (complex III). The electron is transferred from complex III to complex IV through cytochrome C which is a mobile carrier present in the inner membrane. Complex IV is called cytochrome C oxidase complex and consists of cytochromes a-a3, and two copper centers.

Q.114 Distinguish between the following:

(a) Aerobic respiration and Anaerobic respiration

(b) Glycolysis and Fermentation

(c) Glycolysis and Citric acid Cycle

Ans.

(a)

Aerobic respiration Anaerobic respiration
1 It occurs only in the presence of molecular oxygen. It occurs in the absence of molecular oxygen.
2. It is highly efficient and produces 38 ATP molecules. It is less efficient and generates only 2 ATP molecules.
3. It takes place in both cytoplasm and mitochondria. It takes place in the cytoplasm only.
4. It is a multistep process having glycolysis, Krebs’ cycle, and ETS. It is not a multistep process.
5. It produces carbon dioxide and water as the by-product. It produces ethyl alcohol and carbon dioxide as a by-product.

(b)

Glycolysis Fermentation
1. It is a common step in both aerobic and anaerobic respiration. It is strictly an anaerobic mode of respiration.
2. It results in the production of pyruvic acid. It produces ethyl alcohol.
3. Net gain is 8 ATP molecules. Net gain is 2 ATP molecules.
4. The product of glycolysis is used as an intermediate in Krebs’ cycle. The product of fermentation (ethyl alcohol) is not used by cells further.

(c)

Glycolysis Citric acid Cycle
1. It occurs in the cytoplasm. It occurs in the mitochondria.
2. It is a non-cyclic process. It is a cyclic process.
3. It is common in both aerobic and anaerobic respiration. It takes place only in aerobic respiration.
4. It produces 8 ATP molecules from one molecule of glucose. It produces 15 ATP molecule from one molecule of Acetyl CoA.

Q.115 What are the assumptions made during the calculation of net gain of ATP?

Ans.

Many assumptions have been made in order to calculate the net gain of ATP from one molecule of glucose. This is required as the cellular system is very complex where numerous biochemical reactions take place simultaneously. The assumptions are as follows:

  1. All the steps of aerobic respiration (glycolysis, TCA cycle, ETS and oxidative phosphorylation) take place in sequential order where the product of the first step enters the subsequent step as a substrate.
  2. The NADH synthesized in glycolysis enters in the mitochondria and produces ATP through oxidative phosphorylation.
  3. The intermediates formed in various steps of aerobic respiration are not utilized in any other metabolic pathways other than subsequent steps of respiration.
  4. The glucose molecule is the only substrate and no other molecule enters as an intermediate substrate in the process.

Q.116 Discuss “The respiratory pathway is an amphibolic pathway.”

Ans.

The process of metabolism involves both anabolic and catabolic reactions. Anabolism is the synthesis of complex macromolecules like lipids and proteins from simple molecules like glycerol and amino acid respectively. On the other hand, catabolism includes the breakdown of macromolecules into simple molecules so that they can enter in the respiratory pathway as a substrate for the release of energy. If fatty acids are used as a respiratory substrate they are broken down to glycerol and acetyl CoA. Glycerol gets converted to 3-phosphoglyceraldehyde (PGAL) and enters in glycolysis while Acetyl CoA directly enters in Krebs’ cycle. However, when an organism needs to synthesize fatty acids, acetyl CoA is withdrawn from the above-said pathway and is made available for catabolic reaction. Similarly, when proteins are used as a substrate, they are first broken down to amino acid, which in turn, depending on their structure, gets converted into different intermediates of Krebs’ cycle. At the time of need, the same molecules are withdrawn to synthesize new proteins. Most of these reactions are reversible and depending on the requirement, the cell uses the respiratory substrate in the process of anabolism or catabolism. Thus, the respiratory pathway is known as the amphibolic pathway rather than only a catabolic pathway.

Q.117 Define RQ. What is its value for fats?

Ans.

Complete oxidation of substrates during aerobic respiration requires oxygen and apart from energy, carbon dioxide is produced as the by-product. The ratio of the volume of CO2 released to the volume of O2 consumed during complete oxidation of one molecule of a substrate in a given period of time at standard temperature and pressure is called the respiratory quotient (RQ).

For example, during aerobic respiration of one molecule of glucose, 6 molecules of CO2 are released and 6 molecules of O2 are consumed. Thus RQ for glucose is 1.

RQ value for fats: Fats need more oxygen molecule than carbohydrate (glucose) for complete oxidation through aerobic respiration, due to which the value of RQ for fat is always less than 1.

For example, when fatty acid tripalmitin is used as a substrate, 145 molecules of O2 are consumed whereas 102 molecules of CO2 are produced, the RQ value is 0.7.

Q.118 What is oxidative phosphorylation?

Ans.

The metabolic pathway that uses the energy released by the oxidation of nutrients to produce adenosine triphosphate (ATP) is called oxidative phosphorylation. Almost all the forms of life on earth use a range of different nutrients to carry out oxidative phosphorylation to produce the molecule that supplies energy to metabolism i.e. ATP. This is a very efficient process of energy generation.

This process requires the presence of oxygen in the system. Oxygen drives the whole process as it removes hydrogen from the system and acts as the final hydrogen acceptor. During oxidative phosphorylation, electrons are transferred from electron donors like NADH2 to electron acceptors such as oxygen. These redox reactions release energy, which is used to form ATP. In eukaryotes, these redox reactions are carried out by a series of protein complexes within mitochondria, whereas, in prokaryotes, these proteins are located in the cells’ inner membranes. These linked sets of proteins are called electron transport chains. It is the energy of the oxidation-reduction process that is used for the production of proton gradient required for phosphorylation and thus, this process is called oxidative phosphorylation.

Q.119 What is the significance of step-wise release of energy in respiration?

Ans.

Carbohydrates, proteins, fats and organic acids are used as respiratory substrates and oxidation of these compounds releases energy in the cell. However, the energy released is not dissipated freely in the cell. In other words, it does not occur in one step. Instead, it is released in a series of slow step-wise reactions controlled by enzymes and is trapped in the form of ATP. This prevents the sudden increase in the temperature and avoids wastage of energy. This holds a lot of significance as ATP which stores the energy can be broken down whenever and wherever it is needed in the various energy-requiring processes of the organisms.

Q.120 Answer in one word or one line.

(i) Give the common name of Periplanata Americana.

(ii) How many spermathecae are found in earthworm?

(iii) What is the position of ovaries in cockroach?

(iv) How many segments are present in the abdomen of cockroach?

(v) Where do you find Malpighian tubules?

Ans.

(i) Cockroach

(ii) Eight spermathecae are found in earthworm.

(iii) The two large ovaries are found lying laterally in the 2nd-6th abdominal segment.

(iv) 10 segments are present in the abdomen of both male and female cockroaches.

(v) Malpighian tubules are found at the junction of midgut and hindgut of a cockroach.

Q.121 Answer the following:

(i) What is the function of nephridia?

(ii) How many types of nephridia are found in earthworm based on their location?

Ans.

(i) Nephridia (sing.: nephridium) are the excretory organs of earthworm. They regulate the volume and composition of body fluids and excrete out the wastes of the body through a pore to the surface in the body.

(ii) Based on the location, three types of nephridia are found in earthworm:

a) Septal nephridia: They are present on both the sides of intersegmental septa of segment 15 to the last. They open into the intestine.

b) Integumentary nephridia: They are very minute and are hardly visible to the naked eye. They are attached to the lining of the body wall from segment 3 to the last. They open separately on the body surface.

c) Pharyngeal nephridia: They are present as three paired structures at the sides of the oesophagus in 4th, 5th and 6th segment.

Q.122 Draw a labelled diagram of the reproductive organs of an earthworm.

Ans.

The diagram showing the reproductive organs of an earthworm is depicted below:

Q.123 Draw a labelled diagram of alimentary canal of a cockroach.

Ans.

The diagram showing the alimentary canal of a cockroach is depicted below:

Q.124 Distinguish between the followings

(a) Prostomium and peristomium

(b) Septal nephridium and pharyngeal nephridium

Ans.

(a)

Prostomium Peristomium
It is the lobe at the anterior end of earthworm that serves as a covering for the mouth. It is the first body segment of an earthworm. It is also known as buccal segment.
It is sensory in function. It also helps to crack open the soil when earthworm wants to enter in it. It contains the mouth.

(b)

Septal nephridium Pharyngeal nephridium
It is present on both sides of the intersegmental septa of segment 15 to the last one. It is present as three paired tufts in the 4th, 5th and 6th segment.

Q.125 What are the cellular components of blood?

Ans.

Blood is a fluid connective tissue. It contains red blood cells (RBCs), white blood cells (WBCs) and platelets (together constituting ~45% of volume) suspended in plasma (~55% of volume).

Red Blood Cells (erythrocytes): They are biconcave, large microscopic cells without nuclei. They carry oxygen from the lungs to the body’s tissue and take carbon dioxide back to the lungs to be exhaled out from the body.

White Blood Cells (leukocytes)- They exist in variable numbers and types (granulocytes and agranulocytes) but constitute only about 1% of blood’s volume. Leukocytes are not limited to blood. Most are produced in bone marrow from the same kind of stem cells that produce red blood cells.

Platelets (thrombocytes)- They are small, colourless cell fragments without nuclei in the blood. Their main function is to interact with clotting proteins to stop or prevent bleeding.

Q.126 What are the following and where do you find them in animal body.

(a) Chondrocytes

(b) Axons

(c) Ciliated epithelium

Ans.

(a) Chondrocytes are cells of healthy cartilage tissue. They are found in the small cavities within the matrix secreted by them. Examples of cartilage tissue are the tip of the nose, outer ear joints, etc.

(b) The long slender projections of the nerve cells of nervous tissues are called axons. They help in sending messages by conducting electrical impulses away from the neuron’s cell body. They are present in the neural tissues like brain, spinal cord, nerves etc.

(c) When the columnar or cuboidal cells of epithelia bear cilia on their free surface, they are called the ciliated epithelium. They help in the movement of particles or mucus in a particular direction over the epithelium. They are present in the inner surface of hollow organs like bronchioles and fallopian tubes.

Q.127 Distinguish between

(a) Simple epithelium and compound epithelium

(b) Cardiac muscle and striated muscle

(c) Dense regular and dense irregular connective tissues

(d) Adipose and blood tissue

(e) Simple gland and compound gland

Ans.

(a)

Simple epithelium Compound epithelium
It is composed of a single layer of cells. It is composed of two or more layers of cells.
It provides lining for body cavities, ducts and tubes of the body. It provides protection against chemical and mechanical stresses.
It is involved in the process of secretion, diffusion, absorption, etc. Its limited role is secretion and absorption due to the presence of multiple layers.
It is found in the ducts of glands, tubules of nephrons, walls of blood vessels, air sacs of lungs, the lining of alimentary canal etc. It covers the dry skin surface, the moist surface of the buccal cavity, pharynx, salivary glands and pancreatic ducts.

(b)

Cardiac muscle Striated muscle
It is a contractile tissue which has the ability to contract. It is a bundle of fibres grouped together in a parallel fashion.
It is involuntary in nature. It is voluntary in nature.
They are found only in the heart. They are found in biceps, triceps and limbs.

(c)

Dense regular connective tissues Dense irregular connective tissues
Fibres and fibroblasts show a regular orientation in dense regular connective tissues. Collagen fibres are present in rows between many parallel bundles of fibres. Fibres (collagen) and fibroblasts show an irregular orientation in dense irregular connective tissues.
Example: Tendons and ligaments Example: Skin

(d)

Adipose tissue Blood tissue
It is a type of loose connective tissue, where adipocytes and fibres are loosely arranged. It is a type of fluid connective tissue, composed of red blood cells, white blood cells and platelets in plasma.
It is semi-fluid in nature. It is fluid in nature.
The cells are specialised to store fats. The cells help in the transportation of food, wastes, gases and hormones from one part of the body to another. Also involved in defence and coagulation.

(e)

Simple gland Compound gland
It is also known as a unicellular gland. It is also known as a compound (multicellular) gland.
It consists of isolated glandular/secretory cells. It consists of a cluster of glandular/secretory cells.
Example: Goblet cells of the alimentary canal. Example: Salivary gland.

Q.128 Mark the odd one in each series:

(a) Areolar tissue; blood; neuron; tendon

(b) RBC; WBC; platelets; cartilage

(c) Exocrine; endocrine; salivary gland; ligament

(d) Maxilla; mandible; labrum; antennae

(e) Protonema; mesothorax; metathorax; coxa

Ans.

(a) Neuron

(b) Cartilage

(c) Ligament

(d) Antennae

(e) Protonema

Q.129 Match the terms in column I with those in column II:

Column I Column II
(a) Compound epithelium (i) Alimentary canal
(b) Compound eye (ii) Cockroach
(c) Septal nephridia (iii) Skin
(d) Open circulatory system (iv) Mosaic view
(e) Typhlosole (v) Earthworm
(f) Osteocytes (vi) Phallomere
(g) Genitalia (vii) Bone

Ans.

Column I Column II
(a) Compound epithelium (iii) Skin
(b) Compound eye (iv) Mosaic view
(c) Septal nephridia (v) Earthworm
(d) Open circulatory system (ii) Cockroach
(e) Typhlosole (i) Alimentary canal
(f) Osteocytes (vii) Bone
(g) Genitalia (vi) Phallomere

Q.130 Mention briefly about the circulatory system of earthworm.

Ans.

Characteristic features of the circulatory system of earthworm:

  • It consists of blood vessels, capillaries and heart.
  • It carries food, waste and respiratory gases along with coelomic fluid.
  • The blood vascular system is closed in nature and blood remains confined to heart and blood vessels.
  • The contractions of heart pump blood in one direction.
  • Smaller blood vessels supply the gut, nerve cord and the body wall.
  • Blood cells and haemoglobin are produced in blood glands present in 4th, 5th and 6th segment and then dissolved in the blood plasma. Blood cells are phagocytic in nature.

Q.131 Draw a neat diagram of digestive system of frog.

Ans.

The diagram showing the digestive system of a frog is depicted below:

Q.132 Mention the function of the following:

(a) Ureters in frog

(b) Malpighian tubules

(c) Body wall in earthworm

Ans.

(a) The two ureters emerging from the kidneys of male frog act as a urinogenital duct that carry both urine and sperms. In female frogs, ureter carries only urine and excretes it out through cloaca.

(b) Malpighian tubules present at the junction of midgut and hindgut is involved in the removal of excretory products from the haemolymph of cockroaches.

(c) The body wall in earthworm performs the following functions:

  • Helps retain water
  • Helps in motion
  • Helps in respiration (diffusion of gases)
  • Helps in the contraction of the earthworm
  • Secretes mucus which helps in locomotion and burrowing
  • Also performs a sensory function

Q.133 Describe various types of epithelial tissues with the help of labelled diagrams.

Ans.

Epithelial tissue is the tissue provides lining to a body part and has a free surface either facing body fluid or the outside environment. There are two major types of epithelial tissues:

  1. Simple epithelium: It is composed of a single layer of cells and functions as a lining for body cavities, ducts and tubes. On the basis of structural modification, the simple epithelium is further divided into three types
  • Squamous epithelium: Single thin layer of flattened cells with irregular boundaries mostly involved in the diffusion process. E.g. cells of walls of blood vessels, air sacs of lungs.
  • Cuboidal epithelium: Single layer of cube-like cells. They are mostly found in ducts of glands and nephrons and are mainly involved in secretion and absorption.
  • Columnar epithelium: Single layer of tall and slender cells with microvilli on the free surface. They are found in the lining of stomach and intestine and are involved in secretion and absorption.

Sometimes, cuboidal or columnar epithelium bear cilia (known as ciliated epithelium) and move particles in a specific direction or get specialized for secretion (known as glandular epithelium).

  1. Compound epithelium: It consists of two or more cell layers and has protective function e.g. our skin. They mainly provide protection against chemical and mechanical stresses. They form a dry protective layer of the skin, the moist surface of the buccal cavity, pharynx, inner lining of ducts of salivary glands and of pancreatic ducts.

Q.134 What are the factors affecting the rate of diffusion?

Ans.

Diffusion is a process of passive movement (ATP/energy is not required) of solutes from one place (higher concentration) to another (lower concentration) until the solute concentration on both the places is equal. Several factors that affect the rate of diffusion are:

  1. Concentration gradient: Diffusion rate depends upon the concentration of solute at the two points. Larger the concentration difference between the two points, higher is the rate of diffusion.
  2. Permeability of the membrane: More permeable the membrane that is separating the solutes, faster is the process of diffusion.
  3. Temperature: Increase in temperature increases the rate of diffusion.
  4. Distance: Diffusion rate is inversely proportional to the distance solute has to travel.
  5. Size of the molecules: Smaller molecules diffuse faster.

Q.135 What are porins? What role do they play in diffusion?

Ans.

Porins are the proteins that form huge pores in the outer membranes of bacteria, mitochondria and chloroplast. They act as pores by forming channels in the membrane through which a variety of molecules can diffuse. They are large enough to allow passive diffusion of solutes across the membrane. There are different types of porins for transportation of different types of molecules.

Q.136 Describe the role played by protein pumps during active transport in plants.

Ans.

Active transport requires energy to facilitate the transport of molecules against their concentration gradient. It is needed where molecules cannot cross the plasma membrane by simple diffusion or when transport of molecules against their concentration gradient is required. There are specific protein pumps (protein channels) that facilitate the transport of such molecules by using the energy from ATP. Hydrolysis of ATP induces a conformational change in protein pumps that allows movement of molecules through it. When all the protein transporters are in use, the transport reaches its peak level or attains a saturation level. This is a highly selective process.

Q.137 Explain why pure water has the maximum water potential.

Ans.

Water potential refers to the tendency of water molecules to flow from one place to another as a result of gravity, osmosis, or capillary action. It mainly deals with the free energy (kinetic energy in a liquid and gaseous state) of water molecules. Presence of any other molecule in water decreases the kinetic energy of its molecules. In the absence of any other molecules, kinetic energy (free energy) of water molecules will be highest and so will the water potential. Thus, pure water has maximum water potential since it has maximum kinetic energy.

Q.138 Differentiate between the following:

(a) Diffusion and Osmosis

(b) Transpiration and Evaporation

(c) Osmotic Pressure and Osmotic Potential

(d) Imbibition and Diffusion

(e) Apoplast and Symplast pathways of movement of water in plants.

(f) Guttation and Transpiration.

Ans.

(a) Diffusion and Osmosis

Diffusion Osmosis
Diffusion is the passive movement of molecules along a concentration gradient. Osmosis is the diffusion of solvent (water) across a semi-permeable membrane.
It can occur in a liquid or gaseous state. It occurs in liquid medium only.
It does not require the presence of water. Presence of water is a must.
It does not require a semi-permeable membrane. It requires a semi-permeable membrane.

(b) Transpiration and Evaporation

Transpiration Evaporation
It is the loss of water from plants through stomata found on the bottom surface of the leaves. It is the loss of water from any surface (living or non-living).
It is a physiological process. It is a physical process.
It is influenced by many physiological and environmental factors. It is driven by only environmental factors.

(c) Osmotic Pressure and Osmotic Potential

Osmotic Pressure Osmotic Potential
Osmotic pressure is the pressure which needs to be applied to a solution to prevent the inward flow of water across a semi-permeable membrane. Osmotic potential is the numerical value of osmotic pressure with an opposite sign.
It is a positive pressure and its value increases with increase in the concentration of solute particles. It is negative pressure and its value decreases with increase in the concentration of solute.

(d) Imbibition and Diffusion

Imbibition Diffusion
Imbibition is the absorption of water by solids or colloids by simple diffusion. Diffusion is passive movement of molecules, ions or particles from one area to another along their concentration gradient.
It involves the movement of liquid. It involves the movement of liquid, gas and solid particles.

(e) Apoplast and Symplast pathways of movement of water in plants

Apoplast pathways Symplast pathways
Apoplast pathway of movement involves the transport of water through the space found between two adjacent cell walls of epidermis and cortex. It involves the transport of water through protoplast.
It is a faster process and water moves by mass flow. It is a relatively slow process.

(f) Guttation and Transpiration

Guttation Transpiration
It is the loss of water from leaves of plants in the form of water droplets. It is the loss of water from leaves in the form of vapours.
It usually occurs at night or early mornings. It occurs during the day.
It is an uncontrolled process and occurs at the endings of leaf veins. It is a controlled process and occurs through stomata.

Q.139 Briefly describe water potential. What are the factors affecting it?

Ans.

Water potential measures the tendency of water to move from one area to another. Water potential is expressed in potential energy per unit volume and is represented by the Greek letter Ψ. Pure water has maximum water potential and mixing of any solute in it decreases its water potential. The greater the concentration of water in a system, the higher is its kinetic energy or water potential. Water potential of pure water at standard temperature and pressure is zero. Solute potential (the magnitude of lowering of water potential due to dissolution of a solute in water) and pressure potential (the magnitude of increase in water potential due to the application of pressure greater than atmospheric pressure) are the two important components that determine the water potential.

Factors affecting water potential are:

Water potential is affected by solute concentration and pressure.

1. Increase in solute concentration decreases water potential and is known as solute potential.

2. Application of pressure (greater than atmospheric pressure) increases water potential and is known as pressure potential.

Q.140 What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?

Ans.

When the pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It helps in the transport of water in biological systems like in plants. When water enters the cell, it builds pressure against the cell wall and subsequently, the cell wall exerts an equal pressure in the opposite direction to maintain its rigidity.

Q.141 (a) With the help of well-labeled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.

(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.

Ans.

(a) Plasmolysis is the loss of water from cells when they are surrounded by a hypertonic (or more concentrated) solution. During plasmolysis, water moves out of the cell, first from the cytoplasm and then vacuoles. This results in detachment of the cell membrane of the plant cell from the cell wall. In plasmolysis, cytoplasm shrinks and gets detached from the cell wall. The cell is said to be plasmolysed. It is a reversible phenomenon. Thus, during plasmolysis, water movement occurs across the membrane from an area of high water potential to an area of lower water potential, that is, from the cell into its surrounding medium. The shrinkage of the onion peel when placed in the concentrated salt solution is an example of plasmolysis.

(b) Water potential is the tendency of water to move from one place to another. Water flows from the area of high water potential to the area of low water potential. When a plant cell is placed in a solution of high water potential cell, water moves into the plant cell and builds pressure against the wall called turgor pressure. However, the plant cell does not burst because of the presence of the cell wall.

Q.142 How is the mycorrhizal association helpful in absorption of water and minerals in plants?

Ans.

The mycorrhizal association is a symbiotic relationship between plant roots and fungi. Fungal filaments either form a dense network around the plant roots or penetrate them and provide a large surface area (which roots alone cannot provide) for the absorption of water and minerals from the soil surrounding it. This strategy helps in providing adequate amounts of mineral ions and water to the plant.

Q.143 What role does root pressure play in water movement in plants?

Ans.

When minerals enter the roots of the plant, they decrease the water potential inside the root causing water to move in (water moves from an area of high water potential to the area of low water potential). This increases the pressure inside the xylem. This pressure is a positive pressure and called root pressure.

  • It helps in pushing water into the roots up to small heights.
  • It also helps in pushing out excess water in the form of droplets from the tip of grass blades or leaves during the night or early morning.

Q.144 Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?

Ans.

Transpiration is the process of loss of water by mesophyll cells of leaves through stomata. During the day time, stomata are open, so water is lost quickly as it reaches from roots to leaves. This results in the creation of a negative hydrostatic pressure in the mesophyll cells that causes mesophyll cells to draw water from nearby veins of leaves. The negative pressure is gradually transmitted downwards from leaves to roots via xylem tissues. This causes continuous upward movement of the water column. Transport of water in plants is explained by transpiration pull theory. This theory has two essential features:

(i) Cohesion between water molecules and adhesion between water and xylem tissues, and

(ii) Transpiration pull

Factors affecting transpiration are:

(i) Physiological factors like the number and distribution of stomata, and

(ii) Environmental factors like the intensity of light, air, humidity and temperature.

The usefulness of transpiration to plants:

  • Transpiration is helpful to plants as it creates a negative pressure in the xylem that causes continuous upward movement of water.
  • It helps in the absorption of water and minerals from the soil.
  • Water reaching from roots to leaves is used for photosynthesis.
  • Water also provides a cooling effect.
  • It helps in maintaining the turgidity of plant cells.

Q.145 Discuss the factors responsible for ascent of xylem sap in plants.

Ans.

The factors responsible for ascent of sap (transport of water and minerals) are:

  1. Transpiration pull: It creates a negative pressure in the xylem that causes continuous upward movement of sap.
  2. Cohesion (between water molecules) forces and adhesion (between water molecules and xylem tissues) forces along with transpiration pull create a water column within xylem that helps in the continuous upward movement of sap.

Q.146 What essential role does the root endodermis play during mineral absorption in plants?

Ans.

The endodermal cells of the root have proteins embedded in their plasma membrane that actively pump ions from the epidermis into the cytoplasm of the endodermal cells. They let some solutes cross the membrane but not others, thus acting as control points. This helps the plant in controlling the quantity and type of solutes that reach the xylem. Root endodermis allows transportation of ions in one direction only due to the presence of a waxy layer made of the suberin (Casparian Strip). It uptakes the selected mineral ions by active transportation. Minerals present in the soil enter the root epidermis first, either by passive or active transport.

Q.147 Explain why xylem transport is unidirectional and phloem transport bi-directional.

Ans.

Xylem transports water from roots to leaves. Transpiration of water through leaves generates a pull that causes continuous upward movement of water. Thus, transport in the xylem is unidirectional.

Phloem, on the other hand, is the transport vessel for food. Food synthesized by leaves during photosynthesis is transported from leaves to other parts of the plant by phloem (for utilization and storage). During spring, food is transported from storage places to the developing bud. Thus, transport in phloem is bi-directional.

Q.148 Explain pressure flow hypothesis of translocation of sugars in plants.

Ans.

Plants synthesize glucose via photosynthesis that is converted into sucrose for transport through the phloem. Phloem consists of companion cells and sieve tube cells. Pressure flow hypothesis best explains the flow of food in the phloem.

1. Sucrose is transported into a companion cell, then into sieve tube by active transport.
2. Loading of sucrose in phloem cells makes them hypertonic. It causes the movement of water into the phloem cells from adjacent xylem.
3. This increases osmotic pressure in phloem cells and phloem sap moves to the area of low pressure (that is, movement of sap occurs from source to sink along the phloem).
4. At the sink, sucrose is taken out of the phloem cells by active transport. As the amount of sugar decreases, there is a decrease in osmotic pressure that results in the movement of water out of the phloem cells.
5. Removal of water from phloem at the sink again creates a pressure gradient that causes continuous movement of sap from source to sink.

Q.149 What causes the opening and closing of guard cells of stomata during transpiration?

Ans.

Stomata are small openings on leaves. Stomatal pores are surrounded by guard cells which are thick and elastic at the side of pore and thin at the opposite side of the pore. Transpiration and gaseous exchange (exchange of CO2 and O2) takes place through stomata. Stomata are open during the day and closed at night. Turgor pressure plays an important role in the opening and closing of stomata. Whenever there is an increase in turgidity of guard cells, they expand. Because of the thin outer boundary of guard cells, they bulge out, thereby opening the pore of stomata. When turgidity of guard cells decreases, they go back to their normal shape and this results in the closing of stomatal aperture.

Q.150 Briefly describe the structure of the following:
(a) Brain
(b) Eye
(c) Ear

Ans.

(a) Brain: Brain is the main coordinating centre of nervous system in all vertebrates and most invertebrates. It is most complex organ of the body. Brain consists of neurons, glial cells and blood vessels. Brain is made up of three main parts:

  • Forebrain
  • Midbrain
  • Hindbrain

Following diagram shows the structure and location of various parts of human brain.

Forebrain: Forebrain consists of cerebrum, thalamus, and hypothalamus.

  • The cerebrum or cortex is the largest part of the human brain and is a highly folded structure. This is also known as grey matter due to the presence of a large number of neuron cell bodies giving it a greyish appearance. Cerebrum is divided into two halves (hemispheres) by a deep furrow. Two hemispheres are joined to each other with nerve fibres, known as the corpus callosum. The cerebrum is further divided into four lobes: the frontal lobe, parietal lobe, occipital lobe, and temporal lobe. The motor areas, sensory areas and large regions that are neither sensory nor motor in function are known as association areas. These areas are involved in complex functions like inter-sensory associations, memory and communication. The inner part of cerebral hemisphere consists of fibres which are covered with myelin sheath and thus; gives an opaque white appearance, hence it is called white matter.
  • Thalamus is a major coordination centre for sensory and motor signalling. It is involved in functions like thoughts and actions. It is covered by cerebrum.
  • Below thalamus, hypothalamus is present which controls body temperature, hunger and thirst.

Midbrain: The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passes through the midbrain. Four round swellings or lobes constitute the dorsal portion of the midbrain and are known as corpora quadrigemina. It consists of the tectum and tegmentum. It is associated with vision, hearing, motor control, sleep/wake, alertness and temperature regulation.

Hindbrain: The hindbrain is made of the cerebellum, pons and medulla. Cerebellum is a highly convoluted structure and thus; provides extra space for a large number of neurons. Like cerebrum, cerebellum is also divided into two hemispheres. Cerebellum is associated with regulation and coordination of movement, posture, and balance. Pons is made up of fibre tracts interconnecting various regions of the brain. The medulla is connected to the spinal cord and contains regulatory centres for respiration, cardiovascular reflexes and gastric secretions.

(b) Eye: Eyes are organs for the perception of light and colours. The pair of eyes are located in sockets of the skull known as orbits. They are nearly spherical in shape. The wall of the eyeball consists of the following three layers: the outer-most sclera, middle choroid and inner-most retina.

Sclera: It is the outer most layer and consists of dense connective tissues. The anterior portion of this layer is called cornea.

Choroid: it is the middle layer and contains blood vessels. It is bluish in appearance. It thickens at the anterior portion of the eye and forms a ciliary body that holds the lens with the help of other ligaments. The ciliary body further extends to form iris, which gives characteristic colour to the eyes. There is an aperture in the front part of the eyes, called pupil. Pupil is surrounded by iris. Muscles of iris regulate the diameter of pupil.

Retina: Retina is the innermost layer and it contains three layers of cells – from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. Photoreceptor cells are of two types-rods and cones.

Eye is divided into two chambers-

Aqueous chamber: It is the space between the cornea and the lens, and contains a thin watery fluid called aqueous humour.

Vitreous chamber: It is the space between the lens and the retina and is filled with a transparent gel, called the vitreous humour. The place where blood vessels enter and optic nerves leave the eyeball is called blind spot because no photoreceptors are present in this region, so no image is formed at this place. There is another place on the retina where only cones are present and it is fovea. Fovea is the region of highest visual resolution.

(c) Ear: Ears are organ of hearing and maintaining balance. Anatomically an ear is divided into three parts- outer, middle and inner ear.

  • Outer ear: Outer ear consists of the pinna and external auditory meatus (canal). External auditory meatus extends up to the tympanic membrane (eardrum). Pinna and meatus contain sebaceous glands.
  • Middle ear: Middle ear contains three small bones (ossicles) called malleus, incus and stapes. They are attached to each other like a chain. Malleus is attached to tympanic membrane and stapes is connected to inner ear. Middle ear is connected to pharynx by a tube called Eustachian tube. Eustachian tube helps in balancing air pressure at both sides of tympanic membrane.
  • Inner ear: Inner ear is a fluid-filled structure called labyrinth. Labyrinth is divided into two parts- bony and membranous labyrinth. The bony labyrinth is filled with perilymph, while membranous labyrinth is filled with endolymph. Membranous labyrinth is further divided into following parts:
  1. Vestibular apparatus: It is a sac-like structure and is composed of three semi-circular canals and otolith organ (otolith consists of utricule and saccule). Base of the canals are swollen and is called ampulla (ampulla contains crista ampullaris). Otolith contains a projecting ridge called macula. Macula and crista are responsible for maintaining posture and the balance of the body.
  2. Cochlea: Cochlea is a coiled structure and extension of succulus. It contains organ of Corti which acts as the auditory receptor. It is the main organ for hearing.

Q.151 Compare the following:
(a) Central neural system (CNS) and Peripheral neural system (PNS)
(b) Resting potential and action potential
(c) Choroid and retina

Ans.

(a)

Central Nervous System (CNS) Peripheral Nervous System (PNS)
CNS mainly consists of brain and spinal cord. PNS consists of all the nerves (spinal and cranial nerves) associated with CNS.
It has bony protective covering. It is not protected by any bony covering.
It is the main centre for coordination and processing of information. It is not the main coordination centre. It is involved in the conduction of voluntary and involuntary impulses across the body.

(b)

Resting potential Action potential
Resting potential is the potential difference across the membrane of neurons under the resting stage. The membrane at this stage is said to be polarised Action potential is the potential difference across the membrane of neurons when there is the conduction of nerve impulse. The membrane at this stage is said to be depolarised
Membrane becomes more permeable to potassium ions and less permeable to sodium ions. It results in a high concentration of sodium ions outside the membrane. Membrane becomes more permeable to sodium ions resulting in inward movement of sodium ions and outward movement of potassium ions.
The membrane becomes positively charged on the outer side and negatively charge inside. The membrane becomes negatively charged on the outer side and positively charge inside.

(c)

Choroid Retina
Choroid is the middle layer of eyeball. Retina is the innermost layer of eyeball.
It contains numerous blood vessels that supply nutrients and oxygen to retina and other tissues. It contains photoreceptor cells namely cones and rods and is involved in the perception of light and colour.

Q.152 Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre
(b) Depolarisation of the membrane of a nerve fibre
(c) Conduction of a nerve impulse along a nerve fibre
(d) Transmission of a nerve impulse across a chemical synapse

Ans.

(a) Polarisation of the membrane of a nerve fibre: Polarisation of the membrane of neuron occurs during resting phase when there is no conduction of nerve impulse. There are different types of ion channels present on the neural membranes. During the resting phase, membrane is relatively more permeable for potassium ions (K+) and nearly impermeable to sodium ions (Na+). The membrane is also impermeable to negatively charged proteins present inside the cell. As a result, concentration of sodium ions is higher on the outer side of membrane as compared to the inner side. These gradients across the resting membranes are maintained by the active transport of ions by the sodium-potassium pump which transports three Na+ outwards for two K+ into the cell. The outer side of the membrane becomes positively charged and the inner side becomes negatively charged due to a difference in the concentration of ions. This is known as polarisation of the membrane of a nerve fibre.

(b) Depolarisation of the membrane of a nerve fibre: When there is a nerve impulse at the site of a polarised membrane, permeability of membrane towards sodium ion changes. The membrane at the site of impulse becomes freely permeable to sodium ions. This causes a rapid influx of sodium ions resulting in increased sodium ion concentration at the inner side of the membrane. The inner side of the membrane becomes positively charged and the outer side becomes negatively charged. This reversal in polarity of the membrane is called depolarisation.

(c) Conduction of a nerve impulse along a nerve fibre: Conduction of nerve impulse is carried out by depolarisation of plasma membrane of nerve cell. Whenever there is a nerve impulse, it opens the sodium ion channel present on the membrane of nerve fibres resulting in the rapid influx of sodium ions to the inside of the nerve fibre. This generates a positive charge at the inner side of the membrane and a negative charge on outer side. This potential difference is called action potential and the state of nerve fibre is said to be in depolarised state. Just ahead of this point, the polarity of the axon membrane is opposite i.e. the outer side is positive and inner side is negative. This results in flowing of current from site of action potential to further down the axon. On the outer surface, current flows in reverse order completing the circuit. The polarity at the site of origin of impulse is reversed and action potential is generated further ahead. As the nerve impulse moves forward, the membrane becomes depolarised that causes the nerve impulse to travel forward.

(d) Transmission of a nerve impulse across a chemical synapse: Junction between two neurons is called synapse. Nerve impulse travels from one neuron to another through these synapses. At a chemical synapse, the membranes of the two neurons (pre-synaptic and post-synaptic neurons) are separated by a synaptic cleft, which is filled with a fluid. The nerve impulse is transmitted from presynaptic neuron to post-synaptic neuron with the help of neuro-transmitters in the following manner:

  • Axon terminal of a presynaptic neuron contains neurotransmitters in vesicles.
  • When there is nerve impulse, vesicles filled with neurotransmitters fuse with the membrane and release neurotransmitters in the synaptic cleft.
  • At synaptic cleft, neurotransmitters bind to their specific receptors present on the membrane of the post-synaptic neurons.
  • This opens ion channels in post-synaptic neurons resulting in the generation of potential difference.
  • The potential difference leads to the transmission of nerve impulse which may be excitatory or inhibitory from pre-synaptic neurons to post-synaptic neurons.

Q.153 Draw labelled diagrams of the following:
(a) Neuron
(b) Brain
(c) Eye
(d) Ear

Ans.

(a) Neuron:

(b) Brain:

(c) Eye

(d) Ear

Q.154 Write short notes on the following:
(a) Neural coordination
(b) Forebrain
(c) Midbrain
(d) Hindbrain
(e) Retina
(f) Ear ossicles
(g) Cochlea
(h) Organ of Corti
(i) Synapse

Ans.

(a) Neural coordination: Coordination by the neural system is done through electric impulses that send signals to target organs/tissues to act coordinately with other organs or tissues. Neural coordination is performed mainly by brain and neurons in the body. Brain receives the stimulus with the help of nerves and sends signals in the form of an electric impulse to the effector organs.

(b) Forebrain: Forebrain consists of the cerebrum, thalamus, and hypothalamus.

  • The cerebrum or cortex is the largest part of the human brain and is highly folded structure. The cerebrum is divided into two halves (hemispheres) by a deep furrow. Two hemispheres are joined to each other with nerve fibres known as the corpus callosum. The cerebrum is further divided into four lobes: the frontal lobe, parietal lobe, occipital lobe, and temporal lobe. The cerebrum is involved in complex functions like intersensory associations, memory and communication.
  • Thalamus is a major coordination centre for sensory and motor signalling. It is involved in functions like thoughts and actions. It is covered by the cerebrum.
  • Below thalamus, the hypothalamus is present which controls body temperature, hunger and thirst.

(c) Midbrain: It is located between the thalamus of forebrain and pons of the hindbrain. Midbrain consists of the tectum and tegmentum. It is associated with vision, hearing, motor control, sleep/wake, alertness and temperature regulation.

(d) Hindbrain: The hindbrain consists of the cerebellum, pons and medulla oblongata. Like cerebrum, cerebellum is also divided into two hemispheres and is a highly convoluted structure. Cerebellum is associated with regulation and coordination of movement, posture, and balance. Pons connects two hemispheres of the cerebellum. The medulla oblongata is the posterior part of the brain. It is associated with the maintenance of balance and posture of body.

(e) Retina: Retina is the innermost layer of the eyeball and it contains three layers of cells from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. Photoreceptor cells are of two types- rods and cones. Cone cells contain iodopsin pigment that is highly sensitive to bright light. Cone cells are involved in the perception of bright light and colour. Rod cells of retina contain pigment rhodopsin which is highly sensitive to dim light.

(f) Ear ossicles: The middle ear contains three very small bones called ossicles. They are as follows.

  • Malleus: Malleus is attached to the eardrum (tympanic membrane) at one side and to the incus at another side.
  • Incus: Incus is connected with stapes.
  • Stapes: It is attached to the internal ear.

They together transmit sound waves from the external ear to the internal ear.

(g) Cochlea: Cochlea is a coiled structure and extension of the succubus. It contains the organ of Corti which acts as an auditory receptor. It is the main organ for hearing. Cochlea forms three chambers- upper scala vestibule, middle scala media and lower scala tympani.

(h) Organ of Corti: Organ of Corti is the part of cochlea, the internal ear. It is located in cochlear duct between the scala vestibuli and the scala tympani. It is found in mammals only. It contains hair cells and auditory receptors. It is the main site for hearing.

(i) Synapse: In a nervous system, synapse is a gap between the two neurons through which nerve impulse gets transferred from one neuron to another. There are two types of synapse-

(a) If the two neurons are in very close proximity, nerve impulse is transferred directly from one neuron to another like the transfer of nerve impulse within a single neuron. This kind of synapse is called electrical synapse.

(b) If the two neurons are separated by a synaptic cleft, then transfer of nerve impulse is mediated by neurotransmitters and the synapse is called as chemical synapse.

Q.155 Give a brief account of:
(a) Mechanism of synaptic transmission
(b) Mechanism of vision
(c) Mechanism of hearing

Ans.

(a) Mechanism of synaptic transmission: Synaptic transmission is the transfer of nerve impulse from one neuron to another. Synaptic transmission is of two types:

Direct transmission, like at electrical synapse

  • At electrical synapse, the membranes of pre- and post-synaptical neurons have hardly any gap (synaptic cleft is absent).
  • Thus electrical current or the nerve impulse is transferred directly from one neuron to another.
  • This type of transmission is very fast.

Indirect transmission mediated by neurotransmitters, like at chemical synapse

  • At chemical synapse, axon of first neuron (presynaptic neuron) releases neurotransmitters (Acetylcholine) in synaptic cleft (synaptic cleft is a fluid filled gap between axon of first neuron and dendrite of the second neuron).
  • Neurotransmitters then activate the ion channels present on the dendrite of next neuron (called as post-synaptic neuron)
  • This causes the depolarisation resulting in transmission of nerve impulse.
  • This type of transmission is slower as compared to electric transmission.

(b) Mechanism of vision: Retina in the eye balls is the site of light and colour perception. Retina contains photoreceptor cells- rod cells and cone cells.

  • Photoreceptor cells contain light sensitive photopigments that are composed of opsin (a protein) and retinal (an aldehyde of vitamin A).
  • As the light falls on retina passing through pupil, it causes dissociation of retinal from opsin.
  • This causes the change in conformation of opsin resulting in change in membrane permeability in photoreceptor cells and generation of action potential.
  • These action potentials (impulses) are transmitted to the visual cortex area of the brain by the optic nerves,
  • The neural impulses are analysed and image formed on retina is perceived.

(c) Mechanism of hearing

  • External ear receives sound waves and transfers them to the tympanic membrane (eardrum).
  • Sound waves generate vibrations in tympanic membrane which are transferred to the oval window of internal ear through ear ossicles (malleus, incus and stapes).
  • Vibrations are then passed-on to cochlea where they generate waves in lymph.
  • The waves in the lymph induce a ripple in the basilar membrane.
  • It causes the hair cells to bend and press them against the tectorial membrane.
  • As a result, nerve impulses are generated in the associated afferent neuron.
  • These are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

Q.156 Answer briefly:
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina?

Ans.

(a) The retina in the eyeballs is the site of colour perception. Cone cells present in retina recognise colours. There are three types of cone cells which are responsible for the recognition of different wavelength of light (different colours). Cone cells contain light sensitive photo-pigments that are composed of opsin (a protein) and retinal (an aldehyde of vitamin A). As the light falls on cone cells passing through the pupil, it causes dissociation of retinal from opsin. It causes the change in conformation of opsin that results in a change in membrane permeability in photoreceptor cells and generation of an action potential. These action potentials (impulses) are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analyzed and coloured image formed on the retina is perceived.

(b) Vestibular apparatus present in the internal ear is responsible for maintaining the body balance. It is a sac-like structure and is composed of three semicircular canals. Canals contain endolymph (a fluid) and sensors (crista ampullar, present at the base of canals, and macula, present in otolith).

(c) Eyes regulate the amount of light falling on the retina with the help of a pupil. The pupil is an aperture at the front of the eyeballs. It is surrounded by iris that is made up of contractile muscles. Iris contracts or expands to constrict or dilate the pupil which in turn regulates the amount of light passing through it.

Q.157 Explain the following:
(a) Role of Na+ in the generation of action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.

Ans.

(a) The sodium (Na+) ion plays an important role in generating and maintaining the potential difference necessary for the conduction of nerve impulse. When the outer side of the membrane of the nerve fibre is positively charged and the inner side is negatively charged, it is in resting state. At this state, there is a high concentration of sodium ions at the outer side of the membrane of nerve fibres. To generate an action potential, polarity should be reversed i.e. outer side should be negative and inner side of the membrane should be positive. When a neuron gets a nerve impulse, sodium ions cross the membrane of nerve fibres through sodium ion channels. This reverses the charge distribution across the membrane. The outer side of membrane becomes negatively charged, while the inner side becomes positively charged. Potential difference generated by this charge redistribution is called an action potential.

(b) Photoreceptor cells contain light-sensitive photo-pigments that are composed of opsin (a protein) and retinal (an aldehyde of vitamin A). As the light falls on retina passing through the pupil, it causes dissociation of retinal from opsin. It changes the conformation of opsin that results in change in membrane permeability of photoreceptor cells and generates an action potential. These impulses are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analyzed and the image formed on retina is perceived.

(c) The external ear receives sound waves and transfers them to the tympanic membrane (eardrum). Sound waves generate vibrations in the tympanic membrane which are transferred to the oval window of the internal ear through ear ossicles (malleus, incus and stapes). Vibrations are then passed on to cochlea where they generate waves in the lymph. The waves in the lymph induce a ripple in the basilar membrane. It causes the hair cells to bend and presses them against the tectorial membrane. As a result, nerve impulses are generated in the associated afferent neuron that is transmitted via auditory nerves to the temporal lobe of the cerebral cortex of the brain, where the impulses are analysed and the sound is recognised.

Q.158 Differentiate between:
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum

Ans.

(a)

Myelinated axons Non-myelinated axons
These are covered with a myelin sheath. These are not covered with a myelin sheath.
Schwan cells are present in the myelin sheath. Schwan cells are absent.
Node of Ranvier is present. Node of Ranvier is absent.
Conduction of nerve impulse is faster. Conduction of nerve impulse is slower.
Chances of loss of impulse during conduction are less. Chances of loss of impulse during conduction are high.

(b)

Dendrites Axons
Dendrites may be branched or unbranched. Axons are branched structures.
They carry impulse towards the cell body (neuron). They carry impulse away from the cell body.
Myelin sheath is absent. They may or may not have a myelin sheath.
Dendrites are small. They are long and big.
Nissl’s granules are present. They are absent in axons.

(c)

Rods Cones
Rods help to see in dim light. They help to see in bright light.
They are not involved in visualisation of a colour. They help in visualisation of colours.
They contain purple photo-pigment called rhodopsin. They have violet photo-pigment called iodopsin.

(d)

Thalamus Hypothalamus
Thalamus is the part of forebrain. It is a major coordination centre for sensory and motor signalling. Hypothalamus is the part of forebrain that controls body temperature and urge for eating and drinking.

(e)

Cerebrum Cerebellum
It is part of the forebrain. It is part of hindbrain.
It is the centre for controlling activities like intelligence, communication, memory, movement, etc. It controls involuntary activities like balance, body equilibrium, fine movement coordination, sneezing, coughing, etc.
It is located in the anterior portion of the forebrain. It is located just above the brain stem.
It is the largest part of the brain. It is the second-largest part of the brain.

Q.159 Answer the following:
(a) Which part of the ear determines the pitch of a sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock?

Ans.

(a) Cochlea determines the pitch of sound.

(b) Forebrain is the most developed part of brain.

(c) Hypothalamus acts as a master clock of body.

Q.160 The region of the vertebrate eye, where the optic nerve passes out of the retina, is called the
(a) fovea
(b) iris
(c) blind spot
(d) optic chaisma

Ans.

(c) blind spot

[Explanation: Region of the retina where optic nerves pass is called blind spot because this region does not contain any photo-receptors.]

Q.161 Distinguish between:
(a) Afferent neurons and efferent neurons
(b) Impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre
(c) Aqueous humor and vitreous humor
(d) Blind spot and yellow spot
(e) Cranial nerves and spinal nerves

Ans.

(a)

Afferent neurons Efferent neurons
Afferent neurons transmit nerve impulse towards brain or spinal cord. Efferent neurons transmit nerve impulse from brain to effector organs.

(b)

Impulse conduction in a myelinated nerve fibre Impulse conduction in an unmyelinated nerve fibre
In myelinated nerve fibres, Schwann cells form the myelin sheath around the axon. The gaps between two adjacent myelin sheaths are called nodes of Ranvier. In unmyelinated nerve fibres, a single Schwann cell encloses the axon and there is no myelin sheath.
The nerve impulse is transmitted from one node to another. This type of impulse transmission is fast. The nerve impulse is transmitted in a continuous manner along the entire length of the nerve fibre. This type of impulse transmission is slow.

(c)

Aqueous humor Vitreous humor
It is a thin and watery fluid that is present between cornea and lens. It is a transparent gel that is present between the lens and retina.

(d)

Blind spot Yellow spot
Blind spot is the region of retina where optic nerves pass. It is small area on retina that is present at the posterior pole of the eye.
Photoreceptors are absent in this region. Cone cells are present.
It does not sense light due to the absence of photoreceptors. It perceives bright light.

(e)

Cranial nerves Spinal nerves
Nerves arising from brain are called as cranial nerves. There are 12 pairs of cranial nerves. Nerves arising from spinal cord are called as spinal nerves. There are 31 pairs of spinal nerves.

Q.162 Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.

Ans.

Growth- The irreversible permanent increase in the size of an organ or its parts or even of an individual cell brought about by metabolic processes utilizing energy and nutrients over a period of time is called growth.

Differentiation- The process of maturation in which a cell converts into a highly specialized cell by a series of modification in the cell wall and cytoplasm to perform a particular function is called differentiation. For example, to form tracheal cells the cambial cells have to modify their cell wall and also loosen their protoplasm.

Development- The sequential and highly coordinated changes that an organism undergoes throughout their life cycle is called development. For example, in plants development means the changes which take place from seed germination to senescence.

Dedifferentiation- The phenomenon by which a differentiated cell performing specific function regains the capacity of division is termed as dedifferentiation. For example, the formation of meristems – interfascicular cambium and cork cambium from fully differentiated parenchymal cells. It happens only under certain conditions.

Redifferentiation- The process by which dedifferentiated tissues lose the capacity to divide again but mature to perform specific functions is called redifferentiation.

Determinate growth- When the growth of cell/ tissue/organ ceases after a specific size or dimension is attained it is called determinate growth or limited growth.

Meristem- The region in the plant body where actively dividing cells are present is called meristem. They accelerate the growth of plants and these tissues are named according to their location. For example, meristem at the shoot tip is called apical meristem, at the root tip is called root meristem and that in the stem is called lateral meristem.

Growth rate- The increased growth per unit time is termed as growth rate. The growth rate is expressed mathematically in terms of increase in size or number of cell per unit time. If the growth occurs as an increase in total surface area of a cell/ tissue without an increase in the number of cells it is called geometric growth. But when growth is accompanied by an increase in the total number of cells, it is called arithmetic growth.

Q.163 Describe briefly:

(a) Arithmetic growth

(b) Geometric growth

(c) Sigmoid growth curve

(d) Absolute and relative growth rates

Ans.

Arithmetic growth: When a linear curve is obtained by plotting the growth parameter (for example root length) against time, it is called arithmetic growth. A typical example of arithmetic growth is elongation of root at constant rate. Arithmetic growth is the outcome of mitotic division, where one of the two daughter cells continue to divide while other differentiate and mature to perform specific function.

Mathematically, it is expressed as

Lt = L0 + rt

Where, Lt = length at the end of experiment

L0 = Length at the start of experiment

t = Time duration of experiment

r = Growth rate

Geometric growth: Geometric growth results from mitotic division where both the daughter cells retain the capacity to divide. Here, initially the growth is very slow which is called lag phase, but later it becomes very rapid which is called exponential phase. Later on, due to limited nutrient supply, cell division slows down again to attain stationary phase. If growth parameters are plotted against time for geometric growth, a sigmoid curve is obtained.

Sigmoid growth curve: During the early stages of development of plants (say at the time of seed germination), the growth rate is slow, but with time as the cell starts utilizing nutrient, the growth is very rapid and becomes exponential. Later on, when the number of cells increases and the amount of available nutrients become limited, the growth rate becomes stationary. When such growth is plotted against time an S-shaped curve called sigmoid curve is obtained.

Absolute and relative growth rates: Absolute growth rate is the measurement and the comparison of total growth per unit time. The relative growth rate is when the growth is expressed as an increase in specific parameter relative to its initial value per unit time known as relative growth rate.

Q.164 List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/horticultural applications of any one of them.

Ans.

Five main groups of natural plant growth regulators are:

  1. Auxin
  2. Gibberellin
  3. Cytokinins
  4. Ethylene
  5. Abscisic acid

Discovery of Auxin: The first observation of the presence of auxin in plants comes from the experiment of Charles Darwin and Francis Darwin on canary grass. They observed the bending of coleoptiles of canary grass towards the unilateral light source (phototropism). When the tip of the coleoptiles was cut, no bending was reported but when cut coleoptiles tip was placed over agar block it transmitted some chemical to the agar block. On placing this agar block over the cut coleoptiles, it again showed bending towards the unilateral source of light. Later on, in 1926, Auxin was isolated from the tip of coleoptiles of oat seedlings.

The physiological function of auxins: Shoot and root apices are the sites of auxin production in plants from where they are transported to their site of action. The auxins isolated from plants are indole-3-acetic acid (IAA) and indole butyric acid (IBA) while, NAA (naphthalene acetic acid) and 2, 4-D (2, 4-dichlorophenoxyacetic) are artificially synthesised auxins. The major physiological roles played by auxins are as follows:

  • Promote apical dominance in plants
  • Prevent premature falling of fruits and leaves
  • Promotes abscission of older mature leaves and fruits
  • Helps in xylem differentiation
  • Promotes cell division

Agricultural/ horticultural application of auxins are:

  • Used for root induction in cuttings when the plant is propagated through stem cutting
  • Synthetic auxin 2,4-D is used as herbicides that selectively kills the dicotyledonous weeds without harming monocot plants
  • Used for the development of seedless fruits as they promote parthenocarpy in tomatoes
  • Sprayed on plants as they promote flowering for example in pineapples

Discovery of Gibberellins: Japanese farmers reported that few seedlings in rice field grow taller than others and never bears seeds; they called it “bakane” or foolish seedling disease. These seedlings were infected by a fungal pathogen, Gibberalla fujikuroi. Later on, E.Kurosawa, showed the reappearance of symptoms when the sterile filtrate was applied to uninfected plants and the active substance was later identified as gibberellic acid.

The physiological functions of Gibberellins:

  • Help in breaking seed dormancy by activating the group of enzymes, hydrolyses, in the seeds which in turn utilises the stored nutrient.
  • Determine the length of internodes
  • Promote bolting in rosette leaves
  • Delay senescence.

Agricultural/ horticultural application of Gibberellins:

  • Delay fruit senescence, thus fruit remains on the tree for an extended period.
  • Spraying Gibberellins increase the length of grapes stalks.
  • In apple, it leads to elongation and improves the shape of fruits
  • In Sugarcane crop, spraying of it increases the length of internode thus increasing yield.

Discovery of Cytokinins: Cytokinins were discovered by F. Skoog and his co-workers during tissue culture experiment of tobacco stem. They observed that the callus (undifferentiated mass of tissue) differentiate into plant only when it is supplemented with auxin along with coconut milk (or extract from vascular tissue, yeast extract or DNA). Skoog and Miller, latter were able to purify this substance, crystallized it and identified it as a cytokinesis promoting substance. They called it kinetin.

The physiological functions of Cytokinins-

  • Synthesised in the region of rapid cell growth and promote cytokinesis
  • Promote the formation of new leaves
  • Enhance chloroplast formation in leaves
  • Promote lateral shoot growth
  • Delay leaf senescence by enhancing nutrient metabolism

Agricultural/ horticultural application of Cytokinins:

  • Delaying senescence help in long-lasting flower which holds economic importance
  • Differentiation of callus by application of cytokinins has great use in plant tissue culture, thus helps in cloning purpose
  • Prevent apical dominance

Discovery of Ethylene: This is a gaseous hormone produced in large amount by ripening fruits. It was discovered by the observation that when ripen orange is kept with banana, it result in hastened ripening of bananas.

Physiological functions of ethylene:

  • Shows the antagonistic effect of dormancy and break seed and bud dormancy
  • Shows triple response in plants and stimulates shoot and root growth and differentiation
  • Enhances leaf and fruit abscission
  • Induction of femaleness in dioecious flowers
  • Stimulates flower opening
  • Enhances flower and leaf senescence
  • Hastens fruit ripening

Agricultural/ horticultural application of Ethylene:

  • Used to hasten fruit ripening
  • Initiate flowering and synchronises fruit set in pineapples
  • Used as an inducer of female flower in cucumbers thus increasing the fruit yield.

Q.165 What do you understand by photoperiodism and vernalisation? Describe their significance.

Ans.

Photoperiodism: The flowering in certain plants depends not only on the combination of light and dark exposures but also their relative duration. This response is called photoperiodism. It is the ability of the plant to detect and respond to the duration of light (length of day and night). Based on the flowering response of plants toward the length of light condition, they are divided into three classes:

Long Day Plants: These plants flower when the length of day (light duration) exceeds a critical duration, thus, in turn, they need a shorter dark period (night length).

Short Day Plants: For flowering, these plants need day length (light condition) less than a critical duration, thus, in turn, they need a long dark period (night) exceeding a critical duration.

Day-Neutral Plants: Plants in this group do not show any correlation between flowering and duration of light exposure.

Significance: Photoperiodism is a very important phenomenon in the life cycle of a plant as it affects the flowering of the plant. The plant does not flower if it does not receive certain day and light conditions and thereby is not able to complete the life-cycle. Understanding the phenomenon of photoperiodism is highly helpful in horticulture (flowering industry) for cultivating and obtaining flowers throughout the year. This is also an important feature in agriculture. Farmers choose the crop in a given area depending upon the photoperiodism response of the crop.

Vernalisation: The process of initiation of flowering or acquisition of a plants ability to flower in spring by exposure to prolonged cold or low-temperature conditions is called vernalisation. This ensures that reproductive development and seed production occurs in spring and summer, rather than in autumn. In such plants, low temperatures control the flower either in a quantitative or qualitative manner. Several cereal plants such as wheat, barley and rye have two varieties (spring and winter varieties) depending on their requirement for low temperature for flowering and grain filling. A similar phenomenon is observed in biennials herbs such as sugar beet and cabbage, which show vegetative growth in first season and flower and die in the second season following low-temperature exposure. In perennial plants, a period of cold is needed first to induce dormancy and then later, after a certain time frame, plants flower.

Significance: The process of vernalisation ensures that the plant has fully developed vegetative phase and is ready for flowering.

Q.166 Why is abscisic acid also known as stress hormone?

Ans.

Abscisic acid (ABA) is one of the plant growth regulators which helps in increasing the tolerance of plants to withstand stress conditions such as:

  • Water scarcity or high temperature: ABA stimulates the closure of stomata to control the water loss. This makes plant tolerant of such conditions.
  • It favours seed dormancy and inhibits seed germination so that the seeds can withstand desiccation and other environmental conditions unfavourable for growth.

Thus, due to its role in stress tolerance ABA is called the stress hormone.

Q.167 ‘Both growth and differentiation in higher plants are open’. Comment.

Ans.

Both growth and differentiation in higher plants are open. A plant continues to grow throughout its life by adding new shoot, branches, leaves, etc. Growth of a plant is brought about by the meristems located at different locations in the plant. The apical meristem results in the growth of root and shoot apices while lateral meristem increases the girth of the plant. The cells of these meristems have the capacity to divide and form specialized cells that make the plant’s body, while they also self-perpetuate. This type of growth where new cells are always being added to the plant body by the activity of the meristem is called open growth.

The cells derived from root and shoot meristems differentiate and mature to perform specific functions. This process is known as differentiation. In plants, differentiated cells undergo dedifferentitation under certain conditions wherein the cells which had lost the capacity to divide regain the capacity to divide again. During this process, meristems/tissues divide and produce cells that once again lose the capacity to divide but mature to perform a specific function. This process is known as redifferentiation. Thus the differentiation process is also open – cells/tissues arising out of the same meristem differ in the structure at maturity.

Q.168 ‘Both a short day plant and a long day plant can produce can flower simultaneously in a given place’. Explain.

Ans.

The flowering in some plant takes place only when they get light exposure exceeding a critical photoperiod or dark period less than a critical duration; such plants are called long day plants. Similarly, some plants need the less day length or photoperiod and a long dark period exceeding certain critical duration are called short day plants. Both short day plant and long day plant can flower simultaneously in the same place if grown with adequate photoperiods by artificial means. For example, if both long day plant and short day plants are grown under long day condition (say during summer when day are longer) but the short day plants are shifted to dark after a critical photoperiod, then both long day plant and short day plant will flower simultaneously.

Q.169 Which one of the plant growth regulators would you use if you are asked to:
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.

Ans.

(a) Auxin
(b) Ethylene
(c) Cytokinins
(d) Cytokinins
(e) Gibberellic acid
(f) Abscisic acid

Q.170 Would a defoliated plant respond to photoperiodic cycle? Why?

Ans.

A plant where all the leaves are removed is called defoliated plant. A defoliated plant will not respond to photoperiodic cycle. This is because before flowering takes place, the shoot apices have to get modified into flowering apices. Flowering in plants depends on the specific duration of light and dark (photoperiod) which is perceived by leaves. According to the hypothesis, the hormonal substance necessary for flowering is synthesised in leaves in response to specific photoperiod and is transported to shoot apices to induce the formation of flowering apices. In a defoliated plant this hormonal substance is absent due to which they do not respond to photoperiodic cycle.

Q.171 What would be expected to happen if:

(a) GA3 is applied to rice seedlings

(b) dividing cells stop differentiating

(c) a rotten fruit gets mixed with unripe fruits

(d) you forget to add cytokinin to the culture medium.

Ans.

(a) Application of GA3 to rice seedlings results in an increase in the length between two nodes thus increasing the inter-nodal axis which makes the plant tall.

(b) If dividing cells stop differentiating, they form a callus. This may be due to the absence of cytokinin in the system.

(c) If a rotten fruit gets mixed with unripe fruits, it will produce ethylene and enhance the rate of ripening of unripe fruits.

(d) If we forget to add cytokinin to the culture medium, the cell will stop differentiating and form a callus.

Q.172 Which of the following is not correct?

(a) Robert Brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are formed from pre-existing cells.
(d) A unicellular organism carries out its life activities within a single cell.

Ans.

(a) Incorrect

[Note: Robert Brown did not discover the cell. It was Robert Hooke first discovered the cell in 1665.]

Q.173 New cells generate from

(a) bacterial fermentation

(b) regeneration of old cells

(c) pre-existing cells

(d) abiotic materials

Ans.

(c) pre-existing cells

[Note: The cell theory given by Rudolf Virchow in 1855 states that cells divide and are formed from pre-existing cells only.]

Q.174 Match the following

(a) Cristae (i) Flat membranous sacs in stroma
(b) Cisternae (ii) Infoldings in mitochondria
(c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus

Ans.

(a) Cristae (ii) Infoldings in mitochondria
(b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids (i) Flat membranous sacs in the stroma

Q.175 Which of the following is a correct statement:

(a) Cells of all living organisms have a nucleus.
(b) Both animal and plant cells have a well-defined cell wall.
(c) In prokaryotes, there are no membrane-bound organelles.
(d) Cells are formed de novo from abiotic materials.

Ans.

(c) In prokaryotes, there are no membrane-bound organelles.

Q.176 What is a mesosome in a prokaryotic cell? Mention the functions that it performs.

Ans.

Mesosomes are special membranous structures found in prokaryotic cells and are formed by the extension of the plasma membrane into the cell. They are infoldings of the bacterial cell membrane. These extensions are in the form of vesicles, tubules and lamellae.

Functions: They are involved in various cellular processes like:

  1. Cell wall formation during cell division
  2. DNA replication and its separation in daughter cells
  3. Respiration (oxidative phosphorylation)
  4. Secretion by enhancing the surface area
  5. They are rich in enzymes

Q.177 How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?

Ans.

The plasma membrane is composed of lipids that are arranged in a bilayer with their polar head towards outside and the hydrophobic tails towards the inside. Embedded in this lipid bilayer are proteins and carbohydrates. One of the most important functions of the plasma membrane is the transport of molecules across it.

Neutral solutes move across the plasma membrane by the process of simple diffusion along the concentration gradient i.e. from higher concentration to the lower concentration. This is called passive transport and no energy is required for this process to occur. The respiratory gases, oxygen and carbon dioxide diffuse into and out of the cell.

The polar molecules cannot move across the plasma membrane in the same way since the lipid bilayer is nonpolar in nature. Transport of such molecules requires membrane proteins that facilitate transport across the membrane. This is called as facilitated transport. This is mediated by:

a) carrier proteins that facilitate the movement by combining with the molecule to be transported and delivering it to the other side of the membrane after undergoing a conformational change

b

Q.178 Name two cell-organelles that are double membrane-bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.

Ans.

The two double membrane-bound organelles are mitochondria and chloroplasts.

Characteristics of mitochondria:

  • Mitochondria are between 0.5-1 µm in diameter and ~7 µm in length, although the size and shape can vary.
  • Their number varies depending upon the physiological activity of the cells.
  • They are bound by a double membrane with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments. The inner membrane is folded to form structures called cristae, which project into the matrix.
  • The matrix is filled with a gel-like fluid. It contains enzymes that break down carbohydrate-derived products.
  • ATP production occurs at the cristae. The outer membrane forms the continuous limiting boundary of the organelle. The two membranes have their own specific enzymes.
  • Mitochondria contain their own DNA, a few RNA molecules and ribosomes. They can also produce few of their own proteins.
  • They reproduce or divide themselves.

Functions of mitochondria: Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP, hence are called ‘powerhouse of the cell’.

Figure: Structure of Mitochondria

Characteristics of chloroplasts:

  • Chloroplasts are lens-shaped, oval, spherical, discoid or even ribbon-like organelles and are about 4-6 µm in diameter and 1-5 µm in length.
  • Their number varies from 1 per cell of the Chlamydomonas, a green alga to 20-40 per cell in the mesophyll of green leaves.
  • They are mostly found in the mesophyll cells of the leaves.
  • They are bounded by a double membrane. Of the two membranes, the inner membrane is less permeable.
  • The membranes of the thylakoids enclose a space called a lumen.
  • Inside the chloroplasts are found numerous membranes which are arranged into flattened sacs called thylakoids.
  • The thylakoids are piled up like stacks of coins and each stack is called a granum. The flat membranous connections or tubules connecting the thylakoids of various grana are called lamellae. These membranous structures are located in the stroma.
  • Chloroplasts belong to a group of plant organelles known as plastids. Chloroplasts contain chlorophyll and carotenoid pigments.
  • Chloroplasts have their own small, double-stranded circular DNA and ribosomes (70S). They can produce their own proteins.
  • They can also divide to form more chloroplasts. Thus, they resemble photosynthetic prokaryotic organisms.

Functions of chloroplasts:

  • The enzymes required for photosynthesis are located in chloroplasts. The thylakoids contain the green pigment chlorophyll that captures the solar energy required for photosynthesis.
  • The stroma contains all the required enzymes for the synthesis of carbohydrates and proteins.

Figure: Structure of chloroplast

Q.179 What are the characteristics of prokaryotic cells?

Ans.

Characteristics of Prokaryotes:

  • Prokaryotes are a group of organisms that do not have a nucleus and membrane-bound cell organelles.
  • All the prokaryotic cells have a cell wall surrounding the cell membrane.
  • The cytoplasm is the fluid matrix that fills the cell.
  • There is no well-defined nucleus. The genetic material, which can be in the form of a single chromosome or a circular DNA is naked and is not surrounded by any nuclear membrane.
  • Many prokaryotes especially, bacteria have a small circular DNA, known as plasmids, other than the genomic DNA. The plasmid confers certain unique phenotype characters so these bacteria like antibiotic resistance.
  • No membrane-bound organelles are found in prokaryotic cells.
  • Mesosomes, special membranous structures are found in prokaryotic cells that are formed by the extension of the plasma membrane into the cell. They are infoldings of the bacterial cell membrane.
  • The cell envelope of most prokaryotic cells is very complex – it consists of a tightly bound three-layered structure that acts as a protective covering with each layer performing a distinct function. The outermost glycocalyx is followed by the cell wall and then the plasma membrane.
  • Prokaryotes can be motile or non-motile. The motility is provided by flagella.
  • Pili and Fimbriae are also surface structures that do not help in motility but help in attachment of bacteria to some surface.
  • Bacteria, blue-green algae, mycoplasma etc. are examples of prokaryotes.
  • They are generally smaller and multiply rapidly. They vary greatly in shape (bacillus, coccus, vibrio or spirillum) and size.

Q.180 Multicellular organisms have division of labour. Explain.

Ans.

Multicellular organisms are composed of a large number of cells with larger complexity involved in terms of structure and function. These cells vary greatly in size, shape and activities. Some of the examples of various kinds of cells based on the function are: red blood cells are round and biconcave to increase the surface area and nerve cells are long cells as they are required to carry signals over long distances etc. Each kind of cell combines to give rise to tissues, many different tissues organize themselves to form an organ and ultimately the organ system which carries out specific metabolic activities such as respiration, digestion, circulation, excretion etc. Each organ system consists of different organs each of which is assigned specific roles. This shows the division of labour which is needed in a complex body such as that of multicellular organisms. Each cell, tissue, organ and organ system carries out its role depending upon the kind of cell it is made up of.

In simple organisms, like the unicellular Amoeba, which consists of only a single cell, all the functions are carried out within a single cell.

Q.181 Cell is the basic unit of life. Discuss in brief.

Ans.

Cell theory which forms the basis for this statement states that:

  • All living things or organisms are made of cells and their products.
  • New cells are created by old cells through division.
  • Cells are the basic building blocks of life.

All the cells are alive and carry out respiration, reproduction (by the process of mitosis or meiosis) and growth. Cells arise from preexisting cells and become specialized for distinct functions such as; contraction, conduction, secretion, absorption, and protection. All cells have a few things in common e.g. cell membrane, DNA, cytoplasm, and ribosomes which carry out numerous functions required for the activity of the cell. A cell is capable of carrying out all the fundamental activities required to live and thus, is called as a basic unit of life.

Q.182 What are nuclear pores? State their function.

Ans.

Nuclear pores are large protein-lined channels with a complex structure that regulate the transportation of large molecules between the nucleus and the cytoplasm through the nuclear envelope. The nuclear membrane is impermeable to large molecules and thus, safeguards the DNA. In spite of this barrier, nuclear pores allow communication between the nucleus and the cytoplasm.

Functions:

  • Allows small molecules and ions to pass freely, or diffuse, in and out of the nucleus.
  • Nuclear pores allow necessary proteins with specific sequence tags (nuclear localization signals) to enter the nucleus from the cytoplasm.
  • RNA transcribed in the nucleus and proteins that are destined to enter the cytoplasm have nuclear export sequences and are thus released in the cytoplasm through the nuclear pores.

Q.183 Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.

Ans.

The endomembrane is an intercellular system is responsible for the flow of materials from one to another part through vesicles but they are specialised to perform the different functions. Both vacuole and lysosome are its components. Lysosomes are membrane-bound organelles that release hydrolytic digestive enzymes to digest food and also breakdown aged and worn-out cells. This is the reason lysosomes are known as suicidal bags. On the other hand, vacuoles help cells to maintain the shape of the cell. Vacuoles also store food, water and waste products. Vacuoles help in the excretion and osmoregulation in Amoeba.

Q.184
Ans.</strong
(i) Structure of Nucleus: The nucleus is a prominent structure in the eukaryotic cell that can be seen with a light microscope after straining. It acts as a control centre of the cell overseeing the metabolic functioning of the cell as well as characteristics of the cell. The nucleus is composed of the nuclear matrix and nuclear envelope.

Nuclear matrix (nucleoplasm): Nuclear matrix is semifluid in nature and contains the chromatin and the nucleolus. Chromatin is a threadlike material that undergoes coiling or condensation into rod-like structures (chromosomes) just before the cell divides. Chromatin consists of DNA with proteins and some RNA.

One or more spherical bodies called nucleoli are also present in the nuclear matrix. There is no membranous separation between the nucleoli and the rest of the nuclear matrix. This is the site of active ribosomal RNA (rRNA) synthesis and looks darker than the rest of the chromatin under an electron microscope.

Nuclear envelope: The nucleus is separated from the cytoplasm by a double membrane known as the nuclear envelope. These parallel membranes have space between them which is called perinuclear space. It acts as a barrier between the cytoplasm and the nucleus however nuclear pores with a very complex structure allow passage of large molecules (like RNA and proteins). The outer membrane is usually continuous with the endoplasmic reticulum and also bears ribosomes on it.

Diagram of the nucleus:

(ii) Structure of Centrosome: Centrosome is an organelle that usually contains two cylindrical structures called centrioles and serves as the microtubule-organizing centre of the cell. Centrioles are short cylinders with a 9+0 pattern of microtubule triplets which means that each centriole is made up of nine evenly spaced peripheral fibrils of tubulin protein. Each fibril, in turn, is a triplet. The central part of the centriole is also made of proteins and is connected with peripheral triplet tubules by radial spokes (also made of proteins). Amorphous pericentriolar materials surround the two centrioles. The two centrioles of a centrosome are perpendicular to each other.

Function: The centrioles form the basal body of cilia and flagella, and spindle fibres that give rise to spindle apparatus during cell division in animal cells. Before an animal cell divides, the centrioles replicate and the members of each pair are also at right angles to each other.

Diagram of Centrosome:

Q.185 What is a centromere? How does the position of centromere forms the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.

Ans.

Centromere: It is the specialized constricted region of the chromosome where the two sister chromatids remain joined together after replication. There is a disc-shaped structure on the sides of centromere called kinetochore to which spindle fibres attach during cell division.

Classification of chromosomes based on the position of centromeres: Chromosomes are classified into four different groups based on the position of centromeres:

  • Metacentric chromosome: The centromere is present in the middle of the chromosome resulting in two equal arms of the chromosome.
  • Sub-metacentric chromosome: The centromere is present slightly away from the middle position of the chromosome resulting in a longer and a shorter arm of the chromosome.
  • Acrocentric chromosome: The centromere is present close to the end of the chromosome resulting in one extremely short and one very long arm.
  • Telocentric chromosome: The centromere is present at one end of the chromosome.

Q.186 Give the dental formula of human beings.

Ans.

The dental formula represents the number and types of teeth found in half lower and half upper jaw of any species. They are multiplied by two to give the total number of teeth present in that species. Teeth are represented in a specific order of incisor (I), canine (C), premolar (PM) and molar (M). The dental formula of an adult human being is

Q.187 Why are living organisms classified?

Ans.

Classification is a process of grouping things based on easily observable characters. Classification of organisms is done for various scientific reasons. Some of the main reasons are listed below.

  • Classification helps in the identification of specific organisms from the groups of organisms.
  • It makes the study of organisms convenient.
  • It helps in spotting the similarities and differences among the different groups of organisms.
  • It helps in the study of fossils and evolutionary relationships of organisms.

Q.188 Why are the classification systems changing every now and then?

Ans.

Millions of living forms exist on the earth that include plants, animals and microorganisms. Out of them, many are known and have been classified by the scientists under suitable groups, but on the other side, many new species are continuously being discovered. Many of these newly discovered species have a new set of characters and to classify them, either a new system of classification should be developed or an older one should be revised. This brings changes in the existing system of classification every now and then.

Q.189 What different criteria would you choose to classify people that you meet often?

Ans.

We can classify the people we meet often on a various basis such as gender, age range, profession, education, hobbies and common interests.

Q.190 What do we learn from the identification of individuals and populations?

Ans.

In a diverse country like India, we can learn from the identification of individuals and populations about their native places, tribes, food habits, languages and dialects, clothing, religions, castes, customs, etc.

Q.191 Given below is the scientific name of Mango. Identify the correct written name.
Magnifera Indica
Magnifera indica

Ans.

Magnifera indica is the correct scientific name of Mango. It is because in the binomial system of nomenclature, the generic name of the species starts with the capital letter and the specific name starts with a small letter.

Q.192 Define a taxon. Give some examples of taxa at different hierarchical levels.

Ans.

The specific level in the hierarchy of the system of classification of living organisms is called taxon. The following table states the different taxa with an example at different hierarchical levels.

Q.193 Can you identify the correct sequence of taxonomical categories?

(a) Species → Order → Phylum → Kingdom

(b) Genus → Species → Order → Kingdom

(c) Species → Genus → Order → Phylum

Ans.

Both (a) and (c) are the correct sequence of taxonomical categories. The correct sequence of taxonomical categories is Species, Genus, Family, Order, Class, Phylum/Division, and Kingdom.

Q.194 Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meaning of species in case of higher plants and animals on one hand, and bacteria on the other hand.

Ans.

The various definition of species are:

(i) Species is the group of individual organisms with fundamental similarities such as in their morphology, anatomy and physiology.

(ii) Species is the basic unit of classification where individuals share common characteristics.

(iii) Species is the group of individuals who can interbreed among themselves and are capable of producing fertile offspring. They are reproductively isolated from the other members of the different species.

Since higher plants and animals perform sexual reproduction, their species can interbreed freely among themselves to produce the fertile offspring.

On the other hand, bacteria with common genetic material belong to the same species since they perform asexual reproduction.

Q.195 Define and understand the following terms:
(i) Phylum
(ii) Class
(iii) Family
(iv) Order
(v) Genus

Ans.

(i) Phylum: Phylum is a taxon consisting of one or more classes that possess correlated characters. It comes just below the kingdom. In plants, classes with similar characters are assigned under division rather than phylum.

(ii) Class: Class is a group of closely related orders. Class lies above order and below phylum. One example of a class is all members of class Mammalia have mammary glands to feed their offspring and have hair on their bodies.

(iii) Family: Family is composed of different genera with less number of similarities as compared to the genus and species level. It lies above the genus and below order. Example: All tiger, lion and leopard belong to the cat family.

(iv) Order: An order is the taxa with the one or more families of organisms that possess some specifically correlated characters that are different from the members of the different orders. It lies above family and below the class. The similar characters in an order are lesser in number as compared to different genera in the included family.

(v) Genus: Genus is a group of related species which has more common characters in comparison to species of other genera. It lies above species and below family. Example: Lion, leopard and tiger belong to the same genus Panthera.

Q.196 How is a key helpful in the identification and classification of an organism?

Ans.

Key is a taxonomical tool that is made in use for the identification of plants and animals based on the similarities and differences. It is developed on the basis of contrasting characters usually presented in a pair, called couplet, that represents the choice between two opposite characters. The selection of one character in a key leads to the rejection of another character when family, species or genera is identified. For example, the presence or absence of hair is a couplet to identify if the animal is a mammal or not.

Once the family is known, a set of identification key can be used for finding its species and then genus. If the organism is not already recorded, the efforts are made to find its details before giving it a new name.

Q.197 Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.

Ans.

The taxonomical hierarchy of a plant and an animal is given in the following table.

Q.198 Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionisable) functional groups in the amino acids.

Ans.

The titration curves help in identifying the existence of different ionic groups present in the amino acids. In the case of neutral and basic amino acids, the number of dissociating groups is usually one and in acidic amino acids, the number of dissociating groups are usually two.

Q.199 Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?

Ans.

In plants, growth is the symbol of an increase in the quantity of protoplasm. Therefore, measuring the growth of protoplasm can demonstrate the growth of a plant. Like other organisms, plants also grow in various phases of their life cycle. The parameters to measure the growth of protoplasm vary for different parts of the plant; such as the parameters to measure the growth of fruit and seed are very different from each other. Some of the parameters are increased in the height, weight, length, diameter, surface area, volume, and cell number. Measuring growth involves the measurement of the increase of protoplasm in all these parameters. Thus, it is difficult to demonstrate growth throughout the life of a flowering plant using only one parameter.

Q.200 Discuss with your teacher about
(i) Haploid insects and lower plants where cell-division occurs, and
(ii) Some haploid cells in higher plants where cell-division does not occur.

Ans.

(i) Cell division occurs in insects and lower plants, like algae (Spirogyra and Chlamydomonas), bryophytes and pteridophytes. Haploid gametes are produced through mitosis and the zygote formed after fertilisation undergoes meiosis to produce haploid organisms.

(ii) In higher plants, antipodals and synergids of the embryo sac are haploid and do not undergo cell division.

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